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Chapter 7 Answer Key
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0% found this document useful (0 votes)
114 views60 pagesChapter 7
Chapter 7 Answer Key
Uploaded by
Ahmed ElkammarCopyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
/ 60
CHAPTER 7
Trigonometric Identities and Equations
Getting Started, p. 386 3x = 2or 2x = -S
© 8 = - 3x2) Qe 5
1. a) Do the following to isolate and solve for x: 3
Bx 7=5- 9% > 5
Sx - 74 9x oe sage xa >
es «Use the quadratic formula to solve for x as
e-7472547 Phaah
12x = 12 pete
Wax _ 12
x=-lt v2
£) Move all terms to the left side of the equation and
un use the quadratic formula to solve for x as follows:
2 Bx = 3x41
44 32 -3x-1=3rt1-3e-1
xy 3x - 3x -1=0
2
r-45 x
©) Factor the left side of the equation and solve for 32VCy AED
“vas follows. Se
ve - Sx-24=0
(2-8) +3) =0
x-8=Oorr+3=0
x-848=0+8orx43-3=0-3
x= Borx=—3
4) Move all terms to the left side of the equation, 2, To show that AB = CD the distance formula
factor, and solve for x as follows should be used as follows
eT First for AB
s+ ly = 10= 10-1 ~ Vapor
Gx’ + Jlx - 10=0 gh Sse
(Gx = 2) +5) =0
3x-2=0 of 2x+5=0
3x -24+2=042o0r2x +5 ~
Advanced Functions Solutions Manual 7A Now for CD:
d=V(e— nF + O27 nh
y + (6-5
So, AB = CD.
3.aysin a = Opposite side
hypotenuse
cos A = SMiacent side
hypotenuse
opposite side
(an A gjacemt side
exc A = Pupotenuse
“opposite side
hypotenuse
SCA cgiacentside 15°
cot A = Adjacentside _ 15
opposite side 8
b) To determine the measure of 2A in radians, any
ofthe ratios found in part (a) can be used. In this
case, sin A will be used
sina =%
nA ay
1 Se
sin“ (sina) = sino?
8
LA = sin
massa
mZA = 05 radians
72
©) From the figure:
1s
sn B= 35
sin”? (sin B (8)
jin”? (sin BY al)
m2B = sin (8)
mB = 61.9°
4.8) 7,
PS
2,
ene
ao
4-2 Da
4
b) Drawing a segment from (~2,2) down to the
negative x-axis forms a right triangle with two legs
of length 2. The tangent of the related acute angle is
For!
‘The measure in radians of the acute angle that has @
tangent equal to 1 is £50 the vale ofthe elated
acute angle is © radians.
«) Since the point at (—2, 2) is in the second
quadrant, the terminal arm of the angle @ must also
be in the second quadrant. Since the angle with a
terminal arm in the second quadrant and a tangent
3H 3
of ~1 measures * radians, 6 = = radians.
5. a) The x-coordinate of point A is cos 4 while the
y-coordinate is sin =. Therefore, the coordinates of
point A are (*2,%2). The x-coordinate of point B
is cos 5, while the y-coordinate is sin 5. Therefore
8). The
the coordinates of point B are (3
an
x-coordinate of point C is cos *, while the
y-coordinate is sin. Therefore, the coordinates of
point Care (1,4)
‘The x-coordinate of point D the
is cos 7, while the y-coondinae is sin. Therefore,
coorinates of point Dare (~¥,1). The xcoordinate
of point Bs 08, wit te connate sin
‘Therefore, the coordinates of point E are (~
Sm
‘The s-coordinate of point Fis cos *, while the
Chapter 7: Trigonometric Identities and Equations Sn
‘y-coordinate is sin “*, Therefore, the coordinates of
point Fare (4,
an
Gis cos “=, while the y-coordinate is sin
°8). Te xcoordnate of point
3
‘Therefore, the coordinates of point G are (—},
Sn
‘The x-coordinate of point H is cos **, while the
Se
y-coordinate is sin “. Therefore, the coordinates of
cos, while he ycoorinat i sin”. Therefore
the coordinates of point J are (2, -¥#). The
-coordinate of point J is cos, while the
y-coordinate is sin“. Therefore, the coordinates
of point J are (34, ~4).
b) i) cos a is equal to the x-coordinate of the point
(29)
4i) sin“ is equal to the y-coordinate of point J
so cos =
: 4
osinU™ =]
So oye
iii) cos 7 is equal to the x-coordinate of the point at
(=1,0), so cos 7 = =1,
iv) ese a is equal to the reciprocal of the y-coordinate
6. a) Since tan x = ~}, the leg opposite angle x in a
right triangle bas a length of 3, while the leg adjacent
to angle x has a Jength of 4. For this reason, the length
of the bypotenuse can be calculated as follows:
Baga?
9416
25
255
Therefore, if the angle » is in the second quadrant,
the other five trigonometric ratios are as follows:
opposite len _ 3
sin x = CPEORIE 8 _ =.
** hypotenuse 5
cos = Sdiacent ep _ 4
S* "hypotenuse 5
hypotenuse _ 5.
‘opposite leg 3
hypotenuse __5
adjacentleg
‘Advanced Functions Solutions Manual
adjacentleg _ _4
opposite leg 3
Uf the angle xis in the fourth quadrant, the other
five trigonometric ratios areas follows
cotx
sin = PPOs Le _ 3
hypotenuse 5
hypotenuse 5
cxex =
__ hypotenuse _ 5.
Seex® Gdjacentleg 4°
adjacent
con fiacent leg _
opposite leg,
b) To determine the value of the angle x, any ofthe
ratios found in part a) can be used. If in the
second quadrant, sin x = 3, so.x can be solved for as
follows:
sin (sin x)
x=25
If xis in the fourth quadrant, sin x
be solved for as follows’
§, sox can
3
sin“! (sin.x)
x
7.2) tan 6 = 28, cos 0 + O's tue
b) sin? @ + cost @ = 1 is te
©) sec 0 = 51, sin @ # 0 js false, since see is the
reciprocal of cos 6, not sin 0
4) cos’ @ = sin® @ ~ 1is false. This can be shown,
by performing the following operations:
cos’ 6 = sin? @~ 1
cof +1 = sin’ = 141
cos’ @+1= sin? a
cos 6 + 1 — cos @ = sin? @ ~ cos? 8
sin? @ ~ cos? @ = 1
73 Since sin? @ + cos? # = 1is true,
sin? @ ~ cos? @ = 1 must be false,
¢) 1 + tan? @ = sec? @ is true
cos 6
fH cord = S° 5, sing # Ois true.
8, The sinusoidal function y = sin x is transformed
into the function y = asin k(x ~ d) + cby
vertically stretching or compressing the function
) x by a factor of a, horizontally stretching or
are 1
‘compressing it by a factor of reflecting it in the
axis if @ <0, reflecting it in the y-axis is k <0,
vertically translating it units up or down, and
horizontally translating it d units to the right or the
lefi, Therefore, an appropriate flow chart would be
as follows
Perform a vertical stetch/compression by a
factor of ja
aan
vse |}
to determine the horizontal
stretch/compression.
a
Use a and £10 determine whether the function is
reflected in the y-axis or the x-axis,
ieieees ieee
Perform a vertical translation of
units up or down.
aa
| Perform a horizontal translation of d units to
the right or the left
7.1 Exploring Equivalent i
Trigonometric Functions, pp. 392-393
1. a) Answers may vary. For example: The graph is
that of the trigonometric function y = cos 6. Since
the period of y = cos @ is 2ar radians, the graph
repeats itself every 2m radians. Therefore, three
possible equivalent expressions for the graph are
y= cos (@ + 2m), y = cos (# + 477), and
cos (9 ~ 27),
74
b) The trigonometric functions y = cos @ and
y= sin (6 + a) are equivalent. Since the period
of y= sin (+ 2) is 2 rains, the graph repeats
itself every 27 radians. Therefore, three possible
equivalent expressions for he gap sing the sine
tnionarey = (0+ $).~ (0)
and y = sin (0+)
2. a) The graph of the trigonometric function
¥y = esc is symmetric with respect to the origin,
soy = ese is an odd function. Therefore,
ese (= 8) = ~ esc 4. The graph of the trigonometric
function y = seed is symmetric with respect to the
yraxis, so y = sec is an even function. Therefore,
sec (—0) = sec @. The graph of the trigonometric
function y = cot @is symmetric with respect to the
origin, so y = cot @ is an odd function, Therefore,
cot (—8) = ~cot 8.
) If the graph of the trigonometric function
y= ese is reflected in the y-axis, the equation of
the resulting graph is y = ese (—8). Also, if the
graph of y = ese @ is reflected in the x-axis, the
equation of the resulting graph is y = ese 0. Sinee
the graph of y = ese @ is symmetric with respect to
the origin, a reflection in the y-axis is the same as a
reflection in the x-axis. Therefore, the equation
ese (—6) = —cse @ must be true. If the graph of the
trigonometric function y = sec @ is reflected in the
y-axis, the equation of the resulting graph is
y = sec (~8). Since the graph of y = sec 0 is
‘symmetric with respect to the y-axis, a reflection of
the function in the y-axis results in the function itself.
‘Therefore, the equation sec (~8) = sec # must be
‘ve. If the graph of the trigonometric function
Y= cot is reflected in the y-axis, the equation of
the resulting graph is y = cot (~6). Also, if the
raph of y = cot 6 is reflected in the a-axis, the
equation of the resulting graph is y = —cot 8. Since
the graph of y = cot 0 is symmetric with respect to
the origin, a reflection in the y-axis is the same as a
reflection in the x-axis. Therefore, the equation
cot (=8) = ~cot # must be true.
3.a) Since sin 6 = cos G - °),
G-3)-= 5-9
Qn
= cos
6
= cos
Chapter 7: Trigonometric Identities and Equations ‘b) Since cos @ = sin z-0),
oS -sa($-2)
12 12,
gales
ae
¢) Since tan @ = cot
2 Gis G]F Gia Sis
=
4) Since cos @ = sin
ee
on
2
€) Since sin @ =
sin
8
1) Since tan = cot (Z = D
tan =
6
‘Therefore, ese @ = sec
a
Since cos # = sin G i, carol
Therefore, sec = ese @ = )
Advanced Functions Solutions Manual
Since tan @ = cot
Therefore, cot 6 = tan (5 - 0)
>) The trigonometric function y = sec ('
aig)
2) ie
can also be writen as y = se(-(0 ~
iph of the trigonometric function
y= see(-(0~ 3) isteected inte yas, he
: quan ofthe sling graph sy = see (0~ 2
Since the graph of y = see (-(@ ~ 3)) is symmetric
with respect to the y-axis, a reflection of the function
in the y-axis results in the function itself. For this
0) =se(0-2),s0
ee eee
exc @ = see (# ~ 5). Therefore, 5 =e
orsind = cos (0 ~ 2). This isa brown deni,
reason, sec (5
G
the cofunction identity esc @ = sec ( ~ @) must be
twue. The trigonometric function y = ese (F ~ 8) can
5). If the graph
of the trigonometric function y xe (- (0-5
also be writen a8 y = ex (-(0 ~
2)
js reflected in the y-axis, the equation of the resulting
resuling graph sy ~ ce (@ ~ 2), Since he raph
of y= ese(-(0~ 2))isaymmetrie with respect
the origin, a reflection in the y-axis is the same
as a reflection in the x-axis. For this reason,
“ee(§-0)-s=(¢-3)0
os(5-4)-o(0- $4)
se(0+5)
seco
1 1
oi Therefore, > = —7-—y,
sin é =) sec@ ese (0 +
Ts This is a known identity, so the cofunction identity
sec@ = ese (F ~ @) must be tue
‘The trigonometric function y = tan
#) can also
F)). AF the graph of
the trigonometric function y = tan (- (6- 5) is
reflected in the y-axis, the equation of the resulting
araphisy= ton (-(#-2))
Since he saphofy = tan (-(0 ~ 2) isomeric
with respect to the origin, a reflection in the
y-axis is the same as a reflection in the x-axis. For
this eason, tan (2 ~ 6) = an (0~ 2),
tan (¥ - 8) = tan(¢- + 2) = tm(0 +2)
be writen as y = tan (~(0 ~
so cot d= tan (0+ 3), Therefore
1
cord
“aes or tan 6 = cot (@ + 5). This
is a known identity, so the cofunction identity
cot = tan ~ @) most be te
5.) Since sin 8 = sin(m ~ 4),
Tn
8
ee
©) Since tan 6 =
4) Since cos @ =
lin
cos
liz
o> eos(2e - 2%)
16
©) Since sin @
=sin (27 ~ 8),
f) Since tan @ = ~tan (27 ~ 8),
Z
1"
1
g
Is
6. a) Assume the circle is a unit circle, Let the
coordinates of Q be (x, y). Since P and Q are
reflections of each other in the line y = x, the
coordinates of P are (y, x). Draw a line from P to
the positive x-axis. The hypotenuse of the new right
Uwiangle makes an angle of (5 ~ 6) with the
Positive ans, Since the x coordinate of Ps y,
#) = y. Also, since the y-coordinate of
cos
2
Qis y, sin @ = y. Therefore, «os ( 6
'b) Assume the circle is a unit circle. Let the
‘coordinates of the vertex on the circle of the right
triangle in the first quadrant be (x, y). Then
sin @ = y, 50 ~sin @ = ~y. The point on the circle
that results from rotating the vertex by ”
= sing,
‘counterclockwise about the origin has coordinates
(-y.x), 50 cos (F +8) = ~y. Therefore,
oos(¥ +8) = ~sine
7. a) true; the period of cosine is 27
b) false; Answers may vary. For example: Let 8 = 3.
‘Then the left side is sin 5, or 1. The right side is
~sinZ or -1
¢) false; Answers may vary, For example: Let 6 = 7.
Then the left side is eos a, or ~ 1. The right side is
os Sa, oF }
€) false, Answers may vary. For example: Let @ = 2
ea
Then the left side is tan =, or ~Y. The right side
istan 7, or
@) false; Answers may vary. For example: Let @ =
Then the left side is cot, or -1, The right side is
tan 7, oF 1
Chapter 7: Trigonomettic Identities and Equations 1) false; Answers may vary. For example: Let
1. a) Since sin (a + ) = sina cos b + cosasin b,
sin a cos 2a + cos asin 2a = sin (a + 2a) = sin 3a
b) Since cos (a + b) = cos a cos b ~ sin asin b,
0s 4x cos 3x ~ sin 4x sin 3x = cos (Ax + 3x)
= costs
2. a) Since tan (@ ~ b) = 14 tanatans’
tan 170° — tan 110° 7 :
T+ tan 170 tan rie 8 170" 0)
= tan 60° = V3
1b) Since cos (a — b) = cos a cos b + sina sin b,
Sao Sat (= =)
cos == cos + sin sin = = cos (5% = J
2
W 12°" 13 12” 72
eee oe
= cos se = cos =
3. a) Two angles from the special triangles are 30°
and 45°, s0 75° = 30° + 45°.
1b) Two angles from the special triangles are 30° and
45%, s0 — 15° = 30° — 45°.
¢) Two angles from the special triangles are 7 and
g
i or ae
4) Two angles from the special wiangles are and.
pn” 24 6
€) Two angles from the special triangles are 45° and
60°, s0 105° = 45° + 60°
f) ‘Two angles from the special tangles are > and =,
6 6 6 3
4, a) Since sin (a + b) = sin a cos b + cosa sin b,
sin 75° = sin (30° + 45°)
= sin 30° cos 45° + cos 30° sin 45°
OZ) GN
~M2, Mb
oo} 4
= Mi+ v6
~ 4
Advanced Functions Solutions Manual
b) Since cos (a — b) = cos acos b + sin asin b,
20s 15° = cos (45° ~ 30°)
= cos 45° cos 30° + sin 45° sin 30°
“2VE) 20)
ve v3
474
= M+v6
4
tana + tanb
1 tanatanb"
)om(5+)
H+
©) Since tan (a + b) =
on
an a2
are
tan + tan?
3V9+94+343V5
3V3 +3V3-3
12 + 6V3
@) Since sin (a ~ b
w(3)-al
= sin F cos 5 ~ cos 7 sin 5
“AO-A)
Eee
a4
~Mi-ve
4
€) Since cos (a + b) = cosa cosh ~ sina sin b,
‘cos 105° = cos (45° + 60°)
108 45° cos 60) ~ sin 45° sin 60°
“(AQ-AZ) 140
()(0)
_1+0
tana + tanb
tana tan
ea (te) eae
0D 2m i2)7'™
1) Since tan (a + b) =
+em(-2) sat
-n(3) + o()
tana ~ tan
1+ tanatand"
wa):
6 6 6
v3 1
- of) oH)
=04 (4) on
Te
) Since cos (a ~ b) = cosa cos b + sin asin b,
cos m cos = + sin a sin
4 4
we
COPE) OCF) py smecasfa #9) = oes = nasa
cos (+ %) = cos cos ~ sin $ sin Z
= o($)- a
78 Chapter 7: Trigonometric Identities and Equations 6.a) Since sin (a +b) = sina cos b + cosasin b,
sin (a + x) = sin 7 cosx + cos x sinx
(O)(cos x) + (~1)(sin x)
0+ (-sinx)
= sinx
= -sinx
1b) Since cos (a + 6) = cos a cos b ~ sin asin b.
( %) E bs
cos (x +) = c0s x cos
= (cos.x)(0) ~ (sin x)(~1)
= 0 = (Asin x) = sinx
©) Since cos (a + 6) = cosa cos b ~ sinassin b,
cos (x +2) = cosx cos ~ sin x sin
( ) 2
2
= (cos x)(0) = (sin x)(1)
=0-sinx
= -sinx
tana + tanb
€) Since tan (a + b)
1-tanatand’
tanx + tant
ed) 1 ~ tanxtan 7
= tanx
6) Since sin (a ~ b) = sina cos b ~ cosasin b,
sin (x = a7) = sin x cos a ~ cos xsin a7
(sin x)(—1) = (cos .x)(0)
1 tan 2a tan x
__O-tane
1+ (O)(tanx)
Stan
Advanced Functions Solutions Manual
1a) Since sin (a + x) = sin (x + 7), sin (w + 2)
is equivalent to sinx translated 7 units to the left,
which is equivalent 10 ~sin x
by ets (c+ 22) sequen os wasued
3 nits tothe left, which is equivalent to sin x
6 eas (x +2) sequent co tanta
units tothe left, which is equivalent to ~ sin x.
4) tan (x + 7) is equivalent to tan x anslated
fr units to ihe lft, which is equivalent to tan x
2) sin (1 ~ 7) is equivalent 0 sin x translated
x units to the right, which is equivalent to ~sin x.
f) Since tan (24 ~ x) = tan (—x + 2z7), and.
Since the period of the tangent function is 2,
tan (27 = x) is equivalent to tan (~x), which is
squivalent i ton x reflected in the y-axis, which is
>
c+ c-
sin C + sin D = 2sin (SS? eos (S$?) can
be developed as follows:
c+D
vel —
oro) = (m2)
(ode) (22)
in (oeS\u'2)
Wor ae
5) (w$) 102) 2)
- (5) (8)
= sin C + sin D
tana + tan b
T= tanatand’
cot (x + y) in terms of cot x and cot y can be
determined as follows:
cot (x + y)
_ 1
“tan ty) aE
17. Since tan (a + )
1 = tanxtany
van + tany
1 tantany
Chapter 7: Trigonometric Identities and Equations
C cot xcoty
= Solxcoty _ cotxcoty ~ corn coty
To ty eal
corx + cory cot xeoty + cotxcoty
cotzeoty —
__wolseoty _ cot cot y cot x coty
GET CHT ~~ corxeoty ~ cotx + coty
col oot
‘ot x cot
cot + coty
18. Let C= x + yand let D =x ~y.
cos C + cos D = cos (x + y) + cos (x ~ y)
= cos.x cosy ~ sinxsiny
+ cos x cos y + sinx sin y = 2 c08 x c0s y
C+D_xtytx-y
i
: y
7
saensc +0 d= 200 (2) (£52)
19, Let C = x + yand let D =x ~ y.
cos C ~ cos D = cos (x + y) — cos (x = y)
= cos.x cosy ~ sinx sin y
— (cos x cos y — sinx sin y) = —2sinxsiny
C+D xtyix y |
2 2
tae)
2 2 o
So, cos C ~ cos D
7 (& Dyn (S )
ao 2
1. a) Since sin 26 = 2 sin 6 cos 8,
2 sin Sx cos Sx = sin 2(5x) = sin 10x.
b) Since cos 20 = cos? @ — sin® 4,
cos? 8 = sin? 6 = cos 28.
©) Since cos 26 = 1 ~ 2sin*@,
1 ~ 2 sin? 3x = cos 2(3x) = cos 6x.
2tan
4) Since tan 26 = =
er
‘and
(FMB = ta (64) = tne
©) Since sin 26 = 2 sin 0 cos 6,
4 sin 6 cos 6 = (2)(2 sin 8 cos 8
= 2sin 28.
f Since cos 26 = 2 cos? # ~
29 1 ~ cag of 2) =
read? 1 =cn(?) = eose
Advanced Functions Solutions Manual
2. a) Since sin 20 = 2 sin 6 cos 6,
2 sin 45° cos 45° = sin 2(45*) = sin 90°
b) Since cos 26 = cos? @ ~ sin? 8,
1
cos? 30° ~ sin? 30° = eos 2(30°) = cos 60° = >
©) Since sin 28 = 2 sin 6 cos 8.
dino ain(Z) =a =!
4) Since cos 26 = cos? 6 ~ sin® 8.
f) Since sin 26 = 2 sin 8 cos 8,
sin 60°)
cos? 60°"
cos or) 2
= 2 sin 60° cos 60° = sin 2(60°)
2 tan 60° cos? 60° = a
3.a) Since sin 26 = 2 sin # cos 8,
sin 4@ = 2 sin 26 cos 24
b) Since cos 26 = 2cos*# ~ 1
cos 3x = 2 cos*(1.5x) ~ 1
tan
6) Since tan 20 = 7
__2tan (05x)
fans = tan? (058)
4) Since cos 20 = cos — sin? 6,
cos 66 = cos? 39 ~ sin? 30
2sin @ cos 8,
2sin (0.5x) cos (0.5x)
2tan@
T= tan'6
2tan (2.56)
T= tan? (2.58)
1) Since tan 26 =
tan 50 =
4. Since cos @ = 3, the Jeg adjacent to the angle 6
ina right triangle has a length of 3, while the
hypotenuse of the right triangle has a length of 5
For this reason, the other leg of the right triangle
can be calculated as follows,
Bry
75 16
y= 4,in quadrant J
ppOSIE IE sin g
hypotenuse
Therefore, since sin 24 = 2 sin @ cos 6,
moa
Also, since cos 28 = cos? @ ~ sin®
(-G)
Since sin # =
cos 28
paseee
25” 25
ve tan @ = SPPESHE ER on
Finally, since tan @-= STO tan 8 =
2uané
Since tan 20
tan 20
5, Since tan # = —Z, the leg opposite the angle #
in.a right triangle has a length of 7, while the leg
adjacent to the angle @ has a length of 24. For this
reason, the hypotenuse of the right triangle can be
calculated as follows
P+ m=
49+ 576 = 2
¢ = 25, in quadrant I
Since sin @ = a sin 6 = %, and since
adjacent leg
‘nyporenuse? 28 4. (The reason the
sign is negative is because angle @ is in the second
quadrant.) Therefore, since sin 26 = 2sin @ cos @,
— 3. Also, since
cos 8 =
te get
“39 * 537 "527
6. Since sin @ = ~#, the leg opposite the angle 6
in a right triangle has a length of 12, while the
hypotenuse of the right triangle has a length of 13,
For this reason, the other leg of the right triangle
can be calculated as follows:
xiv = 13
xP 144 = 169
x? 4 dd — 144 = 169 ~ 144
25
5, in quadrant IV
Since cos @ cos @ = ¥. Therefore,
byporenuse”
since sin 24 = 2 sin 6 cos 6,
sin 20 = (2)(-#)(,
0s 26 = cos? 6 ~ sin? 8,
opposite leg
adjacent leg’
the sign is negative is because angle @ is in the fourth
quadrant.) Since
2tano
since tan 6 = 42. (The reason
tan 26
tan 20
i-G
7, Since cos 6 the leg adjacent to the angle @
in a right triangle has a length of 4, while the
hypotenuse of the right triangle has a length of 5.
For this reason, the other leg of the right triangle
can be calculated as follows:
3, in quadrant J
‘opposite leg
hypotenuse
Therefore, since sin 26 = 2 sin 6 cos 8,
sin 20 = (2)(3)( 5S
Also, since cos 26 = cos? 0 ~ sin? 6,
Since sin
Chapter 7: Trigonometric Identities and Equations o9-(J-0) we.
6 9 7
25° 25” 25
Finally, since tan 6 = ae tang = ~4) ‘The cosine of 3 is %, so
(he eason the signisnegave sbeause gle is ff ~ eon VD)
in the second quadrant) Since tan 20 = -="25,, Sig ue
8. To determine the value of a, first rearrange the
equation as follows:
2tanx ~ tan 2x + 2a = 1 ~ tan 2x tan? x
2tan x — tan 2x + 2a + tan 2x zy
= tan 2x tan?x + tan 2x * a
Dian +24 1~ tan 2x tan?x + tan2x Since T isin the frst quadrant, the sign of sin Fis
2ianx + 2a~ 1 =1—tan2xtan?x + tan2x~1 positive. Therefore, sin Z = 5H,
2tanx + 2a — 1 = tan 2x ~ tan 2x tan? x e
2tanx + 2a —1= (tan 2x)(1 ~ tan? x) 10. Marion can find the cosine of 5 by using the
2tanx +2a—1_ (tan 2x)(1 ~ tan?x) formula cos 2x = 2 cos? x ~ 1 and isolating cos x
“Tans ante (on one side of the equation as follows:
ae oan aioe cos 2x = 20s? x — 1
(any = y= tan x cos 2x + 1 = 2eos'x-141
2uano cos 2x +1 = 2eos?x
Since tan 20 = j= 8255, the value of 2a —~ 1 must oo TT oe
equal 0. Therefore, a can be solved for as follows: =
2a-1=
+
t-1+1 etsicon2e
2a
2
22
ae
2
9. Jim can find the sine of § by using the formala
cos 2x = 1 ~ 2 sin? x and isolating sin x on one side
of the equation as follows:
cos 2x
cos 2x + 2 sin? x
cos 2x + 2 sin? x
cos 2x + 2sin’ x ~ cos 2x
2sin?x
2sin? x
2
Advanced Functions Solutions Manual 7a Since
is in the first quadrant, the sign of cos =
24+ V3
2
”
is positive. Therefore, cos a
11, a) Since sin 20 = 2 sin 6 cos 0,
sin 4x = 2sin 2x cos 2x
= (2)(2 sin x cos x)(cos 2x). At this point,
either the formula cos 20 = 2 cos" @ — 1,
cos 26 = cos? @ = sin? or cos 20 = 1 ~ 2sin? @
can be used to simplify for cos 2x. If the formula
cos 26 = 2.cos? 6 ~ 1 is used, the formula for sin ax
can be developed as follows:
sin 4x = (2)(2 sin x cos x)(cos 2x)
= (2)(2 sin x cos x)(2.cos? x ~ 1)
= (Asin x c0s 1)(2 cos?.x ~ 1)
= 8cos' xsinr ~ 4sinx cos x
Ifthe formula cos 26 = cos? @ ~ sin? is used, the
formula for sin 4x can be developed as follows:
sin 4x = (2)(2 sinx cos x)(cos 2x)
= (2)(2sin.xc08 x) (cos? x ~ sin? x)
(si ~ sin? x)
= 4 cos’ x sin ~ 4 sin? x cos x
If the formula cos 28 = 1 ~ 2 sin? @ is used, the
formula for sin 4x can be developed as follows:
sin 4x = (2)(2 sin x cos x)(cos 2x)
= (2)(2sinxc0s.x)(1 ~ 2sin®x)
= (4sin x cos x)(1 — 2 sin? x)
= 4 sins cos.x ~ 8sin? x 608 x
b) The valve of sin% is 2. Using the formula
sin 4x = Asin x eos ~ 8sin®x cos x,
3 3 3 . 3
wn = h2)(-3)- (2) (
728
8
sin
sin =
3
12. a) Since sin (a + 6) = sina cos b + cos asin b,
sin 36 = sin (26 + 8) = sin 26 cos @ + cos 26 sin 8.
Since sin 24 = 2 sin 6 cos 8.
sin 39 = (2sin 6 cos #)(cos @) + cos 26 sin
cos? @ sin @ + cos 26 sin 8, At this point, either
the formula cos 20 = 2 cos? @ ~ 1
0s 26 = cos? # ~ sin? 6, or cos 29 = 1 ~ 2sin?@
can be used to simplify for cos 28. If the formula
0s 20 = 2cos’@ ~ 1 is used, the formula for
sin 36 can be developed as follows
sin 39 = 2.cos? @sin @ + cos 20 sin 6
sin 36 = 2 cos? @sin @ + (2 cost @ ~ 1)(sin 8)
sin 39 = 2 cos'@ sin @ + 2.cos? @ sin ~ sin
sin 3 = 4 cos’ @sin @ — sin 8
If the formula cos 24 = cos? @ ~ sin? @ is used, the
formula for sin 36 can be developed as follows:
vial
sin 34 = 2 cos’ @ sin 6 + cos 26 sin 8
sin 30 = 2 cos? @sin 6 + (cos? 6 ~ sin? 6)(sin @)
sin 30 = 2 cos? @sin @ + cos? 0 sin @ ~ sin’ 8
sin 30 = 3 cos? @sin @ — sind 0
If the formula cos 26 = 1 ~ 2 sin? is used, the
formula for sin 38 can be developed as follows:
sin 30 = 2 cos? @ sin 6 + cos 20 sin @
sin 39 = 2 cos’ @ sin @ + (1 ~ 2sin®8)(sin 8)
sin 39 = 2cos*@ sin @ + sin @ ~ 2sin®4
b) Since cos (a + b) = cos.a cos b ~ sin asin b,
cos 39 = cos (26 + 8) = cos 2 cos 6 — sin 6 sin 20,
Since sin 26 = 2.sin 8 cos 8,
cos 34 = cos 28 cos 6 ~ (sin 6)(2 sin 6 cos #)
-08 24 cos 6 — 2 sin? @ cos 8. At this point, either
the formula cos 26 = 2.cos?@ = 1,
cos 20 = cos? @ ~ sin? @, or cos 20 = 1 ~ 2sin?@
can be used to simplify for cos 26. If the formula
cos 20 = 2 cos? 6 = 1 is used, the formula for
cos 39 can be developed as follows
cos 38 = cos 28 cos 8 — 2 sin? # cos
cos 36 = (2 cos" # ~ 1)(cos @) ~ 2sin’ 8 cos @
cos 38 = 2 cos’ 6 ~ cos 4 ~ 2 sin? cos é
If the formula cos 26 = cos’ 0 — sin’ is used, the
formula for cos 39 can be developed as follows:
cos 39 = cos 20-cos 8 ~ 2sin? 6 cos 6
0s 39 = (cos*@ ~ sin? @)(cos 6) ~ 2sin® # cos
608 39 = cos? # = sin? @ cos 6 ~ 2-sin® 8 cos 8
0s 39 = cos’ ~ 3sin? 0 cos #
Chapter 7: Trigonometric Identities and Equations If the formula cos 28 = 1 ~ 2 sin? @ is used, the
formula for cos 39 can be developed as follows:
cos 34 = cos 20 cos # ~ 2sin® # cos 0
cos 38 = (1 ~ 2 sin? @)(cos @) ~ 2 sin? 6 cos 6
cos 36 = cos @ — 2sin’ 6 cos 8 ~ 2sin? 8 cos 6
0s 34 = cos 6 ~ 4sin? @cos 8
tana + tanb
Since tan (a + b
©) Since tan (a+b) = 7 tan a tan B
tan 26 + tan
tan 36 = tan (26 + @) = 0225 Ot
fan 30 = tan (20 + 8) = Tao tan 8
2tan0
Since tan 26 =",
T= tan?
duane
Diane
Stan @ ~ wane
1 tan?
1 tne
_Btang = tant 1 tan?@
T-tan’@ "1 — Stare
_ Stan # = tan’ @
i= Stan'@
13. a) Since sin’ x = §, and since the angle x is in
the second quadrant, sin x = \/§ = 44 = 242,
Since sin x = *¥2, the leg opposite the angle x in a
right triangle has a length of 2/2, while the
hypotenuse of the right triangle has a length of 3
For this reason, the other leg of the right triangle
can be calculated as follows:
xt + (2VOP =F
48-9
xv +8-8
x = —1, in quadrant
[Advanced Functions Solutions Manual
jacent leg
Since cos x = and since the angle 2 is in
sypotenuse
the second quadrant, cos x
Since sin 28 = 2 sin # cos 6.
2 4
sin 2x = 2 sin x c0s.x = aZ\- Nd
by Since sin® x = §, and since the angle x is in the
second quadrant, sin x = \/3 = ¥ =?
Since sin x = *¥2, the leg opposite the angle x in
a right triangle bas a length of 2V3, while the
hypotenuse of the right triangle has a length of 3
For this reason, the other leg of the right triangle
can be calculated as follows:
P+ Qviy= 2
PHB=9
P4+8-8=9
Pel
x= ~1,in quadrant 1
since cos = SUE an since te angle xis
in the second quadrant, cos x = ~3. Since
cos? = sin’ 6,
C3)
(The formulas cos 20 = 2 cos? 6 ~ 1 and cos 20
= 2sin? @ could also have been used.)
8
©) Since sin? x = §, and since the angle x is in the
Vi-4=
Since sin x = 2¥2, the leg opposite the angle x in a
right triangle has a length of 22, while the
hypotenuse of the right triangle has a length of 3
For this reason, the other leg of the right tiangle
can be calculated as follows:
4 QV = 3%
second quadrant, sin x
B+8=9
748-8=9-8
i
= 1, in quadrant I
adjacent leg
Since cos x = SHEER ang since the angle x is,
ypotenvse
in the second quadrant, cos x = —4. Since
°
cos 20 = 2 cos! # — 1, cos @ = 2e0s?5 ~ 1, and
r9 since cos.x = ~4
= Dos = 1-Te vate ot
con® can now be dtm fos
2 Name
4) Since sin? x =
8, and since the angle xis in the
second quadrant, sin.x = \/§
vi
Y= 4 since
sin x = 292, the leg opposite the angle x in right
triangle has a length of 22, while the hypotenuse
of the right triangle has a length of 3. For this
reason, the other ley of the right triangle can be
calculated as follows
H+ Qvay=F
vel
~1, in quadrant
Since cos x = HIMCERSE, and since the angle xis
in the second quadrant, cos x = —}, Since
sin 36 = 3 cos? @ sin 8 ~ sin? 6,
sin 3x = 3 cos? xin x ~ sin?x
a (2) ez)
ai 2) 3 3
of) 194
= OOS cy
6v2 _ 16v2
77
14. If sin a is written as >, where y is the side
opposite angle ain a right triangle, and r is the
radius of the right triangle, the side x adjacent to
7-20
angle a can be found with the formula x? + y? =
Once x is determined, cos a can be written as 3, and
since the terminal arm of angle a lies in the second
quadrant, cos a is negative. With sin a and cos a
known, sin 2a can be found with the formula
sin 26 = 2 sin 6 cos @, Therefore, an appropriate
flow chart would be as follows:
Write sin a in terms of 2.
Solve for x using the Pythagorean
theorem, x? + y?
Choose the negative value of x since
and determine cos a.
Write cos a in terms of 5.
Use the formula sin 2a = 2 sin a cosa
to evaluate sin 2a,
15. a) Since sin 20 = 2 sin 6 cos 0,
sin 2x = 2 sin x cos x. For this reason,
sna tarers
= sin x cos x. Therefore, the
raph of f(x) = sin x cos x isthe same as that of
Als) = 22%. The graph of f(x) = "8% can be
obtained by vertically compressing f(x) = sin x by
a factor of } and horizontally compressing it by a
factor of }. The graph is shown below:
y
b) Since cos 2 = 2 cos? @ ~ 1.
30s 2x = 2 cos? x ~ 1, For this reason,
s 2x + 1 = 2eos?x~ 1 +1 = 2cos*s.
Chapter 7: Trigonometric Identities and Equations ‘Therefore, the graph of f(x) = 2 cos x is the same
as that of f(x) = cos 2x + 1. The graph of
lx) = cos 2x + 1 can be obtained by horizontally
compressing f(x) = cos x by a factor of } and
vertically translating it } unit up. The graph is
shown below:
an 2e
For this reason, a
xy)
Therefore, the graph of
tan 2x
2
Y= tent
> is the same as that of f(x)
Picasa
fo) = 2
‘The graph of f(x) = "2 can be obtained by
vertically compressing f(x) = tan x by
a factor of } and horizontally compressing it by @
factor of }. The graph is shown below:
16. a) To eliminate A from the equations x = tan 2A
and y = tan A to find an equation that relates x10 y,
first take the tan” of both sides of both equations
and solve for A as follows:
x= tan 2A
tan“! x = tan“! (tan 24)
Advanced Functions Solutions Manual
=2A
tan!
ar
= tan A
tan”! (tan A)
tan" y
tan”
and A = tan“ y,
stanly
») To eliminate A from the equations x = cos 24
and y = cos A to find an equation that relates x to y
first take the cos~ of both sides of both equations
and solve for A as follows:
cos 2A
cos” (cos 2A)
=2A
ce
= cosy
) To eliminate. from the equations x = cos 2A
and y = esc A to find an equation that relates x to
first take the cos~? of both sides of the first equation
and the esc"? of both sides of the second equation
and solve for A as follows:
cos 24
cos"! (cos 2A)
2A
csc A
ese" (ese A)
ese“!y
cos? x
and A = cos” y,
cos tx 1
= osc! yor
= sin™
0
y
4) To eliminate A from the equations x = sin 2A
‘and y = see 4A to find an equation that relates x
to y, first take the sin” of both sides of the first
rH equation and the sec” of both sides of the second
equation and solve for A as follows:
x= sin 2A
Since A
sin? x _ sey sin
2 arg 4
17. a) Since cos 2# = 1 ~ 2 sin’ 8, the equation
0s 23
‘This equation can be solved as follows
1 = 2sintx = sin x
1 = 2sin’x + 2sin’x — 1 = sinx + 2sin’x-1
2sintx + sinx 1
(2sinx ~ 1)(sinx + 1) = 0
2sinx-1=0
Qsinx-~1+1=041
2sinx =
2sinx
sinx =
orsinx +1 =
sinxt1~1
sinx
‘Therefore, x
) Since sin 26 = 2 sin 6 cos 6, and since
cos 28 = 2 cos? @ ~ 1, the equation
sin 2x ~ 1 = cos 2x can be rewritten
2sin x eosx — 1 = 2 cos’ x ~ 1. This equation
can be solved as follows:
2sin xcosx ~1 = 2costx-1
2sinxcosx—1+1=2cos?x-1+1
2sinacosx = 2.cos? x
2sin x cos. x ~ 2 sin x cos x
= 2cos? x ~ 2sinx cos x
72
sin x can be rewritten 1 ~ 2 sin? x = sin x.
2eos'x ~ 2sin.x cos.x
(2.008 x)(cos x = sinx
2cosx
2eosx
or cos x ~ sinx = 0
cosx ~ sinx + sinx = 0+ sinx
cosx = sinx
tand
Vi + tan’@
a) sin 26 = 2 sin 6 cos @
Et tan@ (
Vie canta)
2
it
b) cos 2 = cos?@ ~ sin?
y-( tané )
0. V1 + tan’ @,
©) Use the results from parts a) and b)
2rané
Chapter 7: Trigonometric Identities and Equations 2une 4) Since ~cos 6 = cos (m + 6),
i+ tanto
TS tana + 1 ~ ae
1+ tame
2tana
ane
ne
T+ ane
2tan@ | 1+ tan?6
Qn
7
2uané Therefore, ~sin 2 = sin
) Since tan (7 + @) = tan @, tan @ = tan (x + 8),
3a an
2tan’@ 1+ tan?@ ‘Therefore, tan ~~ = tan iG + — )
Beant 4 4
T+ tant@ “Diana ee
= uan6 = ton (4 + =) = tan 2
‘wiachaptar Review pant 2. Since cos 0 = sin (8 +), the equation
1. a) Since cos (2 ~ 6) = cos 6,
cos 6 = cos (2m ~ 8). Therefore,
~6008 (x +2) + 4 canbe rewrten
u ; ~6sin(x +545) +4
cos cos (28 - 2) ~6sin (x + 7) + 4. Since a horizontal
| - translation of a to the left or right is equivalent
to a reflection in the x-axis, the equation
y= ~6sin (x + 7) + 4 can be rewritten
y= 6sinx + 4. Therefore, the equation
y= ~6c0s (x +3) + 4.can be rewritten
yy = 6sinx +4
Therefore, sin“ = sin (= - 3.) Since cos (@ + b) = cos.a.cos b ~ sina sin b,
;
sin (2
oat) 9
©) Since tan (a + 8) = tan 6, tan 6
= tan (108 5 98
a oe
cos (+5) = cos x08 5 ~ sin sin
; os2)(3) = (sin a(- “
1 v
seosx + sinx
2 2
Advanced Functions Solutions Manval 723 1b) Since sin (a + b) sin a cos b + cosa sin b,
sne+ ) = in nfeor®2) + onain®2)
~ na(-“2) + conn(2)
=legsx- 4
= feosx- Bsine
tana + tanb
T= tana tan
Sr
tanx + tan=Z
©) Since tan (a + b) =
1 = tan x tan
tanx +1
wm(e+)-
cam
“Y= (tanx)(Q)
_ 1s tans
“T= tanx
4) Since cos (a + b) = cos a cos b ~ sina sin by
(- ‘) 4a 4a
os (x +5
10s x cos “= — sin x sin =
= toen(2)~ ns
M3 inn
sine = 5
4, a) Since sin (a ~ 6) = sin.a cos b ~ cos a cos b,
in( = uz) = sin x cos 2 — c0s x sin UE
_ 6 ce 6
- «sina -
wot) te)
oe
AS sinx + 5 cos x
2
2
cos x
tana — tan b-
1+ tanatanb
+b) Since tan (a ~ b) =
an (o-f
“7 tanxtan®
3
_ tune V3 tans ~ V9,
“TF an (V5) 1+ Votan
b) = cosa cosh + sinasinb,
(cos (cos 2) + (sin sin 22)
« con( 2) + nn(-°2)
v2
©) Since cos (a
724
) Since sin (a ~ b) ~ sin a cos b ~ cosa sin b,
(sin (cos?) = (cos x) (sn 22)
tsinxy(~4) - (os
fa
1
~ssinx —
2 2
na — tan b
1+ tanatand®
@ — tan 8m Si
g tng an (8 22)
— LE = ton (FZ -
ane = =
* cos x
$.a) Since tan (a ~ b}
= tan = V5
b) Since sin (a ~ b) = sina cos b ~ cos asin b,
299% a 299% m
ee
298-298, 298
sin = 0
©) Since sin (a ~ b) = sin acos b ~ cosasin b,
‘sin 50® cos 20° — cos 50° sin 20°
= sin (50° ~ 20°) = sin 30° = 5
4) Since sin (a +b) = sina cos b + cos asin b,
aa
8
‘
6.8) Since tan a + y= RAINE
2a tanx + tan
tan’x 1 — (tan.x)(tan x)
= tan (x + x) = tan 2x
b) Since sin (a + b) = sin.a cos b + cos asin by
ee ee
sin § cos + cos Fain = sin (5 +
5
= sin = sinx
3s (2 — x) = cos ® cos x + sin sine
dea ($-2)+ ca Zens +0
(0) cos x + (1) sin x = sinx
anan(S +2) =o coer +o
(1) cosx + (0)sin x = cos x
Chapter 7: Trigonometric Identities and Equations oeos(% +2) 4 cos($ +x) = 2005(3 +2)
w oe cons ~ sn a
Geos Fan]
w(B)ae-(Ba]
pun(v-2)= tana tan(@) _ tana
= V3{c0s x ~ sin x)
4)” Te taprian(Z) 1+ tans
1. a= Vaand b = ~3,50
R= VGAP EE
f 1
Vi2 = 2V3
sing
Wi 7
Since cos a is positive and sin a is negative, a is in
the fourth quadrant. @ = =
iien(s+3)
8. a) Since cos 26 = 2 cos? 6 ~ 1,
seo -1=co(ea(2)
So, V3cos x 3 sinx
= cos % = -}
32
b) Since sin 28 = 2 sin 6 cos 4
ia iar ea
sin UE = sin ((2)(=2
asin Feo = so (e0("))
©) Since cos 28 = cos’ ~ sin? 6,
wot ant? = cn (02)
lan In
= cos" = cos
8 8
4) Since cos 24 = 1 ~ 2 sin? a,
. (a(Z))=2=-1
9. a) Since cos! x = Hcos x = +\/B
+92 = MD, and since angle xis inthe
third quadrant, cos x is negative. For this reason,
Mi. Since cos x
cos x =
‘Advanced Functions Solutions Manual
sdjacent to the angle x in a right triangle has a length
of VITO, while the hypotenuse of the right triangle
has a Jength of 11. For this reason, the other leg of
the right triangle can be calculated as follows:
2
(vio) +? =a
mo + y= 121
210+ y? = 110 = 121 = 110
yeu
y = ~ Vii, in quadrant IL
Since sin x = SPP2S228, ang since angle x is in
hypotenuse
the third quadrant, sin.x =
b) Since cos?x = #8, cosx = =}
G35, and since angle x is in the third quadrant,
208 x is negative. For this reason, cos x =
«) Since cos?x = 8. cosx = =V/H
4B, and since angle xis in the third quadrant,
20s x is negative. For this reason, eos x = ~ “il
Since cos x = — 4 the leg adjacent to the angle
rin a right tiangle has a length of VITO, while the
hypotenuse of the right triangle has a length of 11
For this eason, the other leg ofthe right triangle
can be calculated as follows:
(V0 + y? =P
110 + y= 121
110 + y? ~ 110 = 121 ~ 110
ul
‘Vil, in quadrant HT
posite eg
and since angle xis in
wypotenuse
the third quadrant, sin x
sin 26 = 2 sin 0 cos 6, sin 2x
il, since
sin x cos x
= =H, and since angle x is in the third quadrant.
cos x is negative. For this reason, cos x =~“
Since cos x = ~~, the leg adjacent to the angle x
in a right triangle has a length of V110, while the
hypotenuse of the right triangle has @ length of 11
For this reason, the other leg of the right triangle can
be calculated as follows:
(vil0y + y? = 1?
110+ y? = 121
110 + y- 110 = 121 ~ 110
7-25 yeu
y =~ VAT, in quadrant It
opposite leg
Spmerene? and since angle x
is in the third quadrant, sin x = -“2, Since
cos 26 = cos? ~ sin? 6, cos 2x = cos? x ~ sin? x
“BY = B- b= H (The formulas
= 2sin?@
Since sin x =
0s 26 = 2cos? # = 1 and cos 24 =
could also have been used.)
10. Since sin x = 4, the leg opposite the angle x in
a right triangle has a length of 3, while the
hypotenuse of the right triangle has a length of 5.
For this reason, the other leg of the right triangle
can be calculated as follows:
eaves
xe49
vit
:
since sin 26 = 2 sin 6 cos 8,
sin2s~2sinncose ~9(2)(8) 28
Also, since cos 20 = cos? @~ sin? 8,
cos 2x = cos'x — sin? x = (8? - (3)
% (The formulas cos 26 = 2 cos? @ ~
1 ~ 2 sin? @ could also have been used.)
11, Since sin x = {4, the leg opposite the angle x
in a right triangle has a length of 5, while the
hypotenuse of the right triangle has a length of 13.
For this reason, the other leg of the right triangle
can be calculated as follows:
eee
a? + 25 = 169
x? +25 - 25 = 169 ~ 25
v= 1d
x = 12, in quadrant}
Since cos x = *228PUEB gs = H.
Typotenuse
‘Therefore, since sin 26 = 2 sin 6 cos 6,
5\(12) _ 120
sin 2x = 2sin.x cos.x = aX) -3
12. Since cos x = ~4, the leg adjacent to the angle
«xin a right triangle has a length of 4, while the
hypotenuse of the right triangle has a length of 5
7:26
For this reason, the other leg of the right triangle
can be calculated as follows:
e+pas
16+ y? = 25
16 + y?- 16 = 25-16
yao
3, in quadrant 1H
Since tan x = SPPESIEIES tay y = 3, (The reason
aciacent ep
the sign is positive is because angle isin the third
quadrant.) Since tan 29 = 224,
it)
41, Although sin x = cos x is tue for x = 7, it is not
true for all values of x, and therefore, itis not an
identity. A counterexample is a value of x for which
sin x = cos xis not true, Many counterexamples
exist, so answers may vary. One counterexample is
since sin
e 6
2. a) The graphs of f(x) = sin x and
g(x) = tan x cos x are as follows: f(x) = sin x:
AV
WV
tan x cos x:
4, and since cos
6
ax)
) Since the graphs of f(x) = sin x and
a(x) = tan x cos x are the same, sin x = tan x c0s x.
5) To prove that the identity sin.x = tan x c0s x is
rue, tan x cos x can be simplified as follows:
cos x eos x
Chapter 7: Trigonometric Identities and Equations 4 The idensty snot te when cosx = Whecanse
when cosx = 0, tan x oF 2, is undefined. This
Js because 0 cannot be in the denominator of a fraction.
3, a) The graph of y = sin x cot x is as follows:
The graph of y = cos.x is as follows
‘The graph of y
+ 2 sin x cos x is as follows:
Since the graphs are the same, the answer is B.
4) The graph of y = sec? x is as follows:
VU.
Since the graphs are the same, the answer is C
b) The graph o 2ssin! x is as follows:
‘The graph of y = 2 cos? x ~ 1 is as follows
sa 7
NAA
- Pl
Wal)
Since the graphs are the same, the answer is D
€) The graph of y = (sin x + cas.x)? is as follows:
y
Advanced Functions Solutions Manual
oF
‘The graph of. + cos? x + tan? xis as
follows:
is
Sarre Ope ae
Since the graphs are the same, the answer is A.
4, a) The identity sin x cot x = cos x can be proven
as follows
cos x
(sin (2 +)
) The identity 1 ~ 2sin’®x = 2eos? x ~ 1 can be
proven as follows:
1 - 2sin?x = 2cos?x 1
sin.x cos x
sin x cot
cos x
1 2sin?x - 2eos?x 41-0
2 = 2sin?x ~ 2eost'x = 0
2 ~ 2(sin? x + cos? x) = 0
2-20) =0
2-2=0
0=0
«) The identity (sin x + cos x)? = 1 + 2sin x cos.x
san be proven as follows:
(Sin x + cos x)? = sin? x + 2sinx cos.x + cos! x
(sin? x + cos’ x) + 2 sin.x cos x
1+ 2sinxcosx
727 4) The identity sec?x = sin? x + cost x + tan?x
ccan be proven as follows:
sin?x + cost x + tan x
(Sin? x + cos? x) + tan? x
= trans
costx | sin’x
cost x” cos’x
sin? x + cox
sec’ x
5:4) The equation eos x= <2 isnt ne forall
values of x, and therefore, it is not an identity.
A counterexample is a value of x for which
i
cos x = =~ is not tue. Many counterexamples
exist, so answers may vary. One counterexample
2 aw i
is x = % since cos® = ¥, and since —z
6 6 cos =
1a oot .
awyrard
*
b) The equation 1 ~ tan? x = sec? x is not true for
all values of x, and therefore, itis not an identity.
A counterexample isa value of x for which
1 = tan? x= sec* x isnot true, Many counterexamples
exist, so answers may vary. One counterexample
isx = ©, since 1 ~ tan! (7) =1- a?
0, and since sec? () = (Vip =2.
©) The equation
sin (x + y) = cos x cosy + sin x sin y is not true for
all values of x and y, and therefore, itis not an identity.
‘A counterexample is values for x and y for which
sin (x + y) = cos x¢0s y + sin.x sin y is not true
Many counterexamples exist, so answers may vary.
One counterexamples x = and y= a, since
s(n) = sn) = “ham sn
cs(F)om +n)
= (0)(-1) + (YQ)
4) The equation cos 2x = 1 + 2 sin? x is not true
for all values of x, and therefore, it is not an
identity. A counterexample is a value of x for
which cos 2x = 1 + 2sin? x is not true. Many
counterexamples exist, so answers may vary
=1-
7-28
One counterexample is x = 7, since
on (QS) -on (8) “Sato
vase (2-14 a)
cise) fete
6. Answers may vary. For example, the graph of
1 = tax
= is as follows:
TF tants
the funetion y
‘The graph is the same as that of the function
y= cos 2x, So an appropriate conjecture is that
cos 2x is another expression that is equivalent to
1 tani
1 wants
7. The identity fae = cos 2x can be proven as
follows
Ce ——
TFtanx see sect
_ cost = sin?x 1
cor sex
coty—sintx ye F
Oe A x cost x = cost x — sin’ x
cor x
cos 2x
‘tame _ 1 tony
8. The identity 822 can be proven
Foor colx-1
as follows:
1+ tanx = tanx
Ls RS
1+ cotx cotx= 1
_1#tanx 1 = tanx
_lttanx _ d= tanx
= Sanx tT Stan
lan Tans
= tanx = tanx
Since the right side and the left side are equal,
Linx 1 = tanr
Chapter 7: Tigonometric Identities and Equations cost ~ sin 8
9.4) The identity See SEA = 1 ~ tan d can
be proven as follows:
cos? @ ~ sin?
‘cos’ @ + sin 8 cos @
(cos 6 — sin 9)(cos 6 + sin 8)
(cos @)(cos + sin 6)
cosd—sing cod sin@_y ig
cosé cos cos
b) The identity tan? x ~ sin? x = sin? x tan? x can
bbe proven as follows:
LS = tan? x ~ sin? x
‘cos x
w4er,-i)
sin? x(sec? x — 1)
sin? x tan? x
0
sin? x tan? x.
1 + cot? x is a known identity,
Since ese? x
tan? x — sin? x must equal sin® x tan’ x
©) The identity
tan? x = cos? x = 1 ~ cos?x can be proven
cos! x
as follows’
: 1
tan? x — cos*x = —3~ ~ 1 ~ cos?
cos x
2 2 2 1
tan? x — cos x + cos'x = Sad
~ cos? x + cos? x;
tan? x
2
tan? x
tan?x
tan? x
tan? x
1 ae
4) The identity + ph = jean be
proven as follows
1, datos
TF esd * Tred” WF ewsdr~ c0s8)
* T= cos (1 + 0058)
‘Advanced Functions Solutions Manual
_ Le cosd , 1+ 008
“cos 6” 1 costo
_ Le cos +1 + cosé 2
~ 1 cos’ @ 1= cos’ sin? @
10, a) The identity cos x tan’ x = sin x tan? x can
be proven as follows
cos x tan?x = sin x tan? x
cos xtan?x _ sinxtan?x
tan’ x tan? x
cos x tan x = sin x
sinx
cos
coon (3)
sinx
1b) The identity sin’? @ + cos* @ = cos? @ + sin‘ @ can
be proven as follows:
sin? @ + cost @ = cos @ + sint
sin? @ + cost @ — sin’ @ = cos?@ + sin‘ @ ~ sin‘ @
sin? 8 + cost @ ~ sin‘ @ = cos’ 6
sin? @ + cos*@ ~ sin’ @ ~ sin?
cos! @ ~ sin*@ = cos? # ~ sin? @
(cos? @ + sin? 8)(cos? @ ~ sin’ 6)
cos @ + sin? @ = 1
1=1
©) The identity
cot @ ~ sin?
sin x + cos x
(ins + cosn( ee) ee
an tanx sinx
can be proven as follows:
2
tntx+1) 1 2
oe)
tanx ) cosx” sinx
‘sect x cos x
sinx + cosx)(=
i 0sx)| =) x COS X
é _ x ze : )
sinx 1). a= ce
janx)” cosxsinx
cos _ sin x + osx
(sin x + cos x){ x) _ Sinx cose
costx)\sinx) ~~ cos xsin x
arm)
sinx + cos x
cos x sin x
(sin x + cos x)| =
cos x sin x
sinx + cosx
cos x sin x
4) The identity tan? B+ cos? B + sin® B
can be proven as follows
tan? B + cos? 8 + sin? B =
1
cos? B
7:29 tan’ B + 1 = sec’ p
Since tan’ B + 1 = sec! Bis 2 known identity
tan? B + cos! 6 + sin? 8 must equal >. aE
©) The identity
= 4 x) + sin(Z
sin(E +x) + sin(2
can be proven as follows:
we (Es) vaa(2 ~2) = Views
Z cosx + cos % sin x + sin = cos
sin 7 cos.x + cos © sin x + sin cos x
4 4 4
2) = Vicosx
~ cos T sin x = VE.c0s x;
2sin™® con x = VBcos
al¥ 2) (082) Vicos x
Vi cosx = V2 cos x
1) The ident sin (2 — x) con( +) = -sinx
can be proven as follows
sin(Z~sJeai(F +s
=sin x,
a
7
(sn Zens asso)
cos % cos.x ~ sin sin x
x (2
‘af
Sn Zeose teu tans)
(ay (eosx) ~ (0)(sin.x)
((Qiers (sin x
(1){608 x) ¥ (O)(sin x) ~
~sinx = -sinx
11. a) The identity == = cot x can be proven
as follows,
cos 2x + '
sin 2x
=14+1
= cotx
in x 608 x
2oos?x
= cotx
Zsin x cos x
cos x
= cotx
cotx = cotx
b) The identity
2 7 cot x can be proven as
follows
= cotx
cos 2x
2sinx cox og
T= (= 2sin?x)
2sinxeosx
1+ 2si
2sin x cos x
ESR = cot x
2sin?x
cos x
ee cotx
sinx
cot x = cotx
©) The identity (sin x + cos.x)? = 1 + sin 2x can
be proven as follows:
(sinx + cos x? = 1 + sin 2x;
sin?x + sinxcos.x + sinxcos.x + costx
1+ 2sinxcos.x,
sin? x + 2sin x cos.x + cos®x = 1 + 2sin x cos.x;
(cos? x + sin? x) + 2sin x cos
= 1+ 2sin.x cos x;
1+ 2sinxcosx = 1+ 2sinxcosx
4) The identity cos* # ~ sin @ = cos 26 can be
proven as follows:
cos @ ~ sint @ = cos 28
(cos? # + sin? @)(cos? @ ~ sin® 8) = cos? @ = sin? 6
(1)(cos? 6 ~ sin? 8) = cos? @ ~ sin?
cos? @ = sin? 6 = cos? ~ sin? @
€) The idemtity cot @ ~ tan @ = 2 cot 28 can be
proven as follows:
cot @ = tan 6 = 2 cot 26
cos _ sind _ cos 20
m6 cos@ “sin 20
cox sito a cos 20
sinB cos6 cos sin@ 2 eos sind
cos? = sin? @ cos 20
SinBeos@ cos sin
cos 20
cos Osin@ cos@ sind
Chapter 7: Trigonometric identities and Equations
( 1 The identity cot @ + tan 6 = 2esc 24
can be proven as follows:
cot é + tan# = 2ese 20
cos sin@_ 4 1
sin6 cos sin 26
cos? si? olse 1 )
Sin @ cos 8 ” cos @sin@ eos 6 sin 6
cos? @ + sin’ @ 7
sin6cos@ cos sind
1 1
cos #sin@ cos @siné
1+ ane
4) The entity 22 = n(x + 2) canbe
proven as follows:
Tttanx (yo 2
a
tanx + tan =
+tanx __tanx +1
Stanx T= (tanxy()
+tanx 1+ tanx
T—tanx 1—tanx
hh) The idendty ese 2x + eu 2x = cot eau be
proven as follows:
esc 2x + cot 2x
1
sin dy * tan 2x
1
piston)
T—tanx 1 tanxtan®
1
1
i
‘Ot X;
Ol:
sot ae cots
2sinxcosx | _2tanx
tans
L= tan?
eo ee
Tsinxeosx ”” 2tanx
1 tan? x
eet EE - cot x;
2sinxcosx 958%
2sin.x cosx sin sin x”
1, (eosx)(1 = tan? x)(c0s x)
2sin x cos
(cos x)(2 cos.x)
© (sin x)(2 cos x)" :
1, (cos? x)(1 ~ tan! x)
2sin x cos x
2sin x cos x 2sin x cos x
2eostx
nx cosx’
fl cost x
2sin x cos x 2 sin.x cos x
Advanced Functions Solutions Manual
_ 20s? x
* Fein x cos’
1 cos? x
2sinxcosx 2sinxcosx — 2sinxcosx’
Lt cos?x—sin'x __ 2cos?x
2sinxeosx 2 sin x cosx’
Lt costx-sin?x 2eostx
2sinxcosx 2sinxcosx
2cos! x 20s? x
Zsinxcosx 2sin x cosx
1+ cos? x ~ sin? x ~ 2 cos? x
2sinx cosx
1 = sin? x — cos? x
2 sin x cos x
in? x + cos’ x)
2sin x cos x
ini
2sin x cosx
we
2sin x cos x
o=0
2ta9 x indy
can be proven as follows:
Bian x _ sin oe
sin 2x
sin 2x
(2tan x)(cos? x) = sin 2x
sin 2x
sin 2x = 2sin.x cos x
Since sin 2x = 2 sin x cos x is a known identity,
{EEL must equal sin 2x
jp The identity see 2r = SS
can be proven as follows
eset
see 2t
1
cos 21
cos 2
731 2sin?
1 sine
ne
sint “1 = 2sin?r
2sin?t
oe
cos2t cos 2
k) The identity ese 26 = j sec @ csc
can be proven as follows:
esc 26 Fsec@esea
1 ae ) 1
wa (3)(aca)(ana)
Heme
sin 26 cos # sin @
;
2sin@cos@ 2sin6cos6
sin2r_ cos2r
1 The identity see + = S22
can be proven as follows
1__2sinteost
‘The graph is the same as that of the function
y= tam x, $0 an expression equivalent to
sins + snd i. ay
¥ cosx 4 cosy 8 140%
sinx + sin2x
13, The identity S22 S823 = tan x can be
proven as follows:
__ sina + sin 2
1+ cosx + cos 2x
x + 2sinxcos.x
= tanx.
1 + cos x + cos 2x
(sin x)(1_+ 2.c0s x)
= tanx
T+ osx + cos 2x > '"*
x)(1 + 208 x)
T+ cosx + 2eosx-1 “™*
(sin s)(1 + 20x) _
cos x + 20s x
(sinx)(1 + 2 cos x) _
(cos x)(1 + 2cosx) ™"*
aoe = tanx
cos x
tan x = tan
14, A trigonometric identity is a statement of the.
equivalence of twa trigonometric expressions. To
cost Sint prove it, both sides of the equation must be shown
sint __ 2sinteos'r to be equivalent through graphing or simplifying’
costsint sintcosi cos sin rewriting. Therefore, the chart can be completed as
sint__ 2sintcos's 2cos?rsins—sinr _‘follows:
costsini sin reost cos sin Definition Methods of Proof
sint__ 2sin eos’ - (2cos*rsint ~ sins) | A statement of the Both sides of the
cos tsinz sin cost equivalence of two | equation must be
sint __ 2sinscos't ~ 2sinrcos'e + sint trigonometric shown to be equivalent
earTaianie Gress? expressions through graphing or
'
cosfsin’ — cosisint
12. Answers may vary. For example, the graph of
the function y = is as follows
T+ cose + cosde
| simplilying/rewriting.
Trigonometrie
Identities
Examples
cos 2x + sin? x = cos?
cos 2x + 1 = 2cos* x
Non-Examples
cos 2x ~ 2sin? x
cot x + eschx = 1
15. She can determine whether the equation
2sim x cos x = cos 2x is an identity by uying to
simplify and/or rewrite the left side of the equation
so that itis equivalent to the right side of the
Chapter 7: Trigonometric Identities and Equations equation. Alternatively, she can graph the functions
y = 2sin x cosx and y = cos 2x and see if the
graphs are the same. if they're the same, it's an
identity, but if they're not the same, i's not an
identity, By doing this she can determine it’s not
an identity, but she can make it an identity by
changing the equation to 2 sin x cos x = sin 2x.
16. a) To write the expression 2 cos? x + 4 sin x cos x
in the form asin 2x + b cos 2x + ¢, rewrite the
expression as follows:
2 cos? x + 4sin x cos x
Deos? x + (2)(2 sin x c0s x)
(2cos?x ~ 1) + (2)(2sin x cos x) +1
= cos 2x + 2sin 2x +1
= 2sin 2x + cos 2x +1
Since the expression can be rewritten as
2 sin 2x + cos 2x + 1, the values of a, b, and ¢ are
a=2,b=Lande=1
b) To write the expression 2 sin.x cos x ~ 4 sin? x
in the form a sin 2x + b cos 2x + ¢, rewrite the
expression as follows:
~2sinx cosx ~ 4 si
=2sin.x cos x + (2 ~ 4sin?x) ~ 2
~2sinx cos x + (2)(1 ~ 2sin? x) —
~sin 2x + 2 cos 2x ~ 2
Since the expression can be rewritten as
=sin 2x + 2 cos 2x ~ 2, the values of a, b, and c
area = —1,b = 2, and e = ~2.
17. To write the expression 8 cos* x in the form
cos dx + bcos 2x + ¢; it’s first necessary to develop
a formula for cos 4x. Since cos 2x = 2 cos?x ~ 1,
cos 4x = 2.cos? 2x — 1. The formula
cos 2x = 2.cos? x ~ 1 can now be used again to
further develop the formula for cos 4x as follows:
cos 4x = 2 cos? 2x ~ 1
cos dx = (2)(2.c08 x ~ 1)? = 1
cos 4x = (2)(4 cost x ~ 2 cos? x
=2eoext1)~1
cos 4x = (2)(4 cost x = 4eos?x +1) —1
cos 4x = 8cos' x — Scov?x +21
cos dx = Boosts ~ Scostx +1
Since cos 4x = 8 cos‘ x ~ 8cos?x + 1, the value of
@ must be 1. Now, to turn the expression
Scos* x — 8cos? x + 1 into the expression 8 cos* x,
the terms ~8 cos? x and 1 must be eliminated
First, 10 eliminate ~8 cos? x, 4 cos 2x, or
Advanced Functions Solutions Manual
(4)(2.cos?.x — 1), can be added to the expression
8cos* x — 8cos?x + 1 as follows:
Bcostx — Scos?x + 1 + (4)(2cos’x ~ 1)
Bcostx — 8cos?x +1 + 8castx - 4
Bcostx 3
Since adding 4 cos 2x to the expression eliminated
che term ~8 cos® x, the value of b must be 4. Now,
10 turn the expression 8 cos* x — 3 into the expression
8 cos* x, 3 must be added to it as follows:
Beostx -3 +3
Boos’ x
Therefore, the value of ¢ must be 3, and the
expression must be cos 4x + 4 cos 2x + 3.
ibeae=3
78 Solv ava
sin 6, the equation
1a) From the 5 ofy
sin @ = 1 is true when 6
b) From the graph of y = sin @, the equation
Sr
sin @ = ~1is true when
¢) From the graph of y
sin 6 = 0.5 is tue when @
2
sin 6, the equation
Se
ons
4) From the graph of y = sin 6, the equation
on ile
sing Forte
¢) From the graph of y = sin 6, the equation
sin @ = Ois true when 0 = 0, 7, or 27.
£) From the graph of y = sin 8, the equation
= 0.5 is true when
sin @ = is true when @ = 5
2. a) From the graph of y = €0s 6, the equation
cos 6 = 1 is true when 0 = 0 or 277,
) From the graph of y = cos 8, the equation
cos @ = —1 is true when
©) From the graph of y = ¢0s 6, the equation
an
cos @ = 0.5 is true when @ = 3 or.
4) From the graph of y = cos 6, the equation
cos @ = —0.5 is true when 0 = 2 on
©) From the graph of y = cos 8, the equation
cos @ = (is true when 6
f) From the graph of y = cos @, the equation
Ln
cos 8 = “fis true when 6 = F or ~
3. a) Given sin x
733 2 solutions must be possible, since sin x =“
Jn 2 of the 4 quadrants.
b) Given sin. x = “f and 0 = x = 2m, the
solutions for x must occur in the Ist and 2nd
quadrants, since the sine function is positive in
these quadrants,
©) Given sin x = “and 0 = x = 2rr, the related
aes ange for the euaion must be x =, since
4, a) Given cos x = =0,8667 and 0° = x = 360°,
‘wo solutions must be possible, since
cos x = ~0,8667 in two of the four quadrants
b) Given cos x = -0.8667 and 0° = x = 360°, the
solutions for x must occur in the second and third
‘quadrants, since the cosine function is negative in
these quadrants,
©) Given cos x =
related acute angle for the equation must be x
since cos 30° = ~ 0.866.
d) Given cos x = ~ 0.8667 and 0° = x = 360°, the
solutions to the equation must be x = 150°
or 210°, since cos 150° = ~ 0.866 and
cos 210° = - 0.866.
8. a) Given tan @ = 2.7553 and 0 = @ = 277, two
0.8667 and 0° = x = 360°, the
30°,
solutions must be possible, since tan 6 = 2.7553
in two of the four quadrants,
b) Given tan 6 = 2.7553 and 0 = @ = 27. the
solutions for @ must occur in the first and third
‘quadranis, since the tangent function is positive in
these quadrants
¢) Given tan @ = 2.7553 and 0 = @ = 2m, the
related acute angle for the equation must be
1.22, since tan 1.22 = 2.755
4) Given tan @ = 2.7553 und 0 = 6 = 2a the
solutions to the equation must be 6 = 1.22 or 4.36.
since tan 1.22 = 2.7553 and tan 4.36 = 2.7553.
6.) Since tan % = 1 and tan °™ = 1 the solutions
Sa
to the equation tan 6 = 1 are @ = 7 orf
b) Since sin and sin = = J, the solutions
#3
solations tothe equation sin # = dare 9 = or =
734
un
jince cos = = 98 and
©) Since cos € = *f and cos “¢
4) Since sin =
solutions to the equation sin @ = ~
Sm
or.
©) Since cos @? = ~ 4 and cos
solutions to the equation cos 8
Se
or
V3, the
solutions to the equation tan 6 = V3 are @ =
an
f) Since tan 5 = V3 and tan
3
7. a) The equation 2 sin @ = ~1 can be rewritten as
follows:
2sin@ = -1
ano
2° 2
1
2
sin@ =
Given sin @ = —} and 0° = # = 360°, the solutions
to the equation must be @ = 210° or 330°, since
sin 210° = ~} and sin 330° = —}.
) The equation 3 cos @ = ~2 can be rewritten as
follows:
3cos
cos _
2
2
3 3
us
3
cos = —
Given cos @ = 2 and 0° = 6 = 360°, the solutions
to the equation must be @ = 131.8° or 228.2°
€) The equation 2 tan @ = 3 can be rewritten as
follows
2tand =3
Diane _3
22
3
tan 6 = 5
Given tan 6 = § and 0° 160" the solutions to
the equation must be @ = $6.3° or 236.3"
Chapter 7: Trigonometric Identities and Equations @) The equation ~3 sin @ ~ 1
as follows:
can be rewritten
~3sin@~1=1
—3sing-141
~3sine
-3sin@ 2
=3 73
sind = 2
Given sin @ = ~3 and 0° = 6 = 360°, the solutions
to the equation must be 0 = 221.8° or 318.2"
©) The equation —5 cos 9 + 3 = 2 can be rewritten
as follows
~Scos@ +3=2
=5c0sd +3-3=2-
Given cos @ = } and 0° = # = 360°, the solutions to
the equation must be @ = 78.5° or 281.5°
f) The equation § ~ tan @ = 10 can be rewritten as
follows:
8— tend
0
8 — tang + tan 6 = 10+ tan
B= 10 + tand
8-10 = 10+ tan ~ 10
tan@ = ~2
Given tan 8 = 2 and 0° = 6 = 360°, the solutions
to the equation must be @ = 116.6° or 296.6°.
8. a) The equation 3 sin.x = sinx + 1 can be
rewritten as follows
3sinx=sinx +1
3sin x — sinx = sinx +1 ~ sinx
2sinx = 1
2sinx 1
22
1
2
Given sin. x = } and 0 = x = 27, the solutions to
the equation must be x = 0.52 or 2.62
b) The equation 5 cos x ~ V3 = 3 cos x can be
rewritten as follows:
Scosx — V3 = 3cosx
Scosx ~ V3 — 3cosx = 3cosx ~ 3cosx
2eosx ~ V3=0
2eosx — V3+ V3=0+4 V3
Advanced Functions Solutions Manual
2eosx = V3
2eosr V3
2 2
Ma
cos x
Given cos x = %# and
to the equation must be x
¢) The equation cos x ~ 1
rewritten as follows:
cosx 1
cos x — 1 + cos x
2oosx-1=0
2eosx-1+1=041
2cos x
2.cos x
2
2 = 2m, the solutions
52.01 5.76,
cos x can be
~cosx
cos.x + c0sx
cosx =
Given cos x = } and 0 = x = 27, the solutions to
the equation must be x = 1.05 or 5.24.
4) The equation 5 sin x + 1 = 3sinx can be
rewritten as follows
Ssinx +1 = 3sinx
Ssina +1 —3sinx ~3sinx ~3sinx
2sinx +1=0
Qsnx+1-1=0-1
2sinx = -1
2sinx 1
2° 2
ef
sinx = 5
Given sin x = —] and 0 = x = 2n, the solutions to
the equation must be x = 3.67 oF 5.76.
9. a) The equation 2 ~ 2 cot x = 0 can be rewritten
as follows’
2-2eotx=0
2-2eotx + 2cotx = 0+ 2cotx
2 =2eorr
2 _ 2cotx
272
cotx=1
Given cotx = 1 and 0 = x = 2a, the solutions t0
the equation must be x = 0.79 or 3.93.
b) The equation ese x ~ 2 = O can be rewritten as
follows:
escx-2=0
escx-2+2=0+2
cscx = 2
738 Given ese x = 2 and 0 = Seed x 5
d= 143.76
Since the first day when the temperature is
approximately 32°C is day 144, and since the
period of the function is 365 days, the other day
when the temperature is approximately 32°C is
365 — 144 = 221. Therefore, the temperature is,
approximately 32°C or above from about day 144 to
about day 221, and those are the days of the year
‘when the air conditioners are running at the City Hall.
12, When graphed, the function modelling the
height of the nail above the surface of the water is
as follows:
Time(s)
it's necessary to find the first time when the
nail is at the surface of the water. This can be done
as follows:
= ~4sin 201) +25
0-25 = —4sin P(r — 1) +25 - 25
25 = -4sin (1-1)
=25_-4 7
Spat Fe-y
0.625 = sin F0~ 1)
sin“ (0.625) = sin (sin 2 - »)
os7st = 20-1)
0.6751 x4 Fa-1) xt
1-1 = 0.8596
Advanced Functions Solutions Manual
1-141 = 0.8596 +1
1= 1.86
Since the first time when the nail is at the surface of the
water is 1.86 s, and since the period of the function is
85, the next time the nail is at the surface of the water
is6 — 1.86 = 4.145. Therefore, the nail is below the
water when 1.868 <1-< 4.14 5 Since the cycle
repeals itself two more times in the first 24 s that the
‘wheel is rotating, the nail is also below the water when
9.86 s <1 < 12.14 and when 17.868 <1 < 20.145.
13, To solve sin (x +7) = VZeos 2 for
0 =< x = 2m, graph the functions y = sin (x + 4)
and y = V2 cos x on the same coordinate grid as
follows.
Since the graphs intersect when x
Se
x= 5%, the solution to the equation is x
4, Given sin 26 =
agul + Dak or Dk Tree th
%, First, the region d < 3 can be tested by
snding H(2) 8 lows
d)-s00(8)-
since (2) is posite, he Might of the ing
hills above sea level relative to Natasha's home is
positive in the region d < =. Next, the region
3