THE CHINESE UNIVERSITY OF HONG KONG

DEPARTMENT OF MATHEMATICS
MATH1510 Calculus for Engineers (2020-2021)
Solution to Supplementary Exercise 1

Trigonometry

Change of Units

1. Fill in the blanks.

Ans:
π
15◦ = rad;
12
π
30◦ = rad;
6
π
45◦ = rad;
4
π
60◦ = rad;
3
π
90◦ = rad;
2
2π
120◦ = rad;
3
5π
150◦ = rad;
6
180◦ = π rad;
3π
270◦ = rad;
2
360◦ = 2π rad .

Trigonometric Identities

2. Find tan 75◦ and express your answer in surd form.

√
tan 30◦ + tan 45◦ 3
+1 √
Ans: tan 75◦ = tan(30◦ + 45◦ ) = = 3
√ = 2+ 3.
1 − tan 30◦ tan 45◦ 1 − 33
3. Find cos 165◦ and sin 165◦ and express your answers in surd form.
√
1 + 3
Ans: cos 165◦ = cos(120◦ + 45◦ ) = cos 120◦ cos 45◦ − sin 120◦ sin 45◦ = − √ .
2 2
√
3−1
Similarly sin 165◦ = sin(120◦ + 45◦ ) = sin 120◦ cos 45◦ + cos 120◦ sin 45◦ = √ .
2 2
7π 7π
4. Find cos2 and sin2 and express your answers in surd form.
12 12
 2

√ √
7π π π  π π π π 2− 6
Ans: cos = cos + = cos cos − sin sin =
12 3 4 3 4 3 4 4
√ √
7π 2− 3 7π 7π 2+ 3
⇒ cos2 = and sin2 = 1 − cos2 = .
12 4 12 12 4
5. By using the product to sum formula, express each of the following expressions as
a sum of trigonometric functions.
(a) cos 5x cos 3x;
1
Ans: cos 5x cos 3x = (cos 2x + cos 8x).
2
(b) sin 4x sin 2x;
1
Ans: sin 4x sin 2x = (cos 2x − cos 6x).
2
(c) sin 7x cos 3x.
1
Ans: sin 7x cos 3x = (sin 4x + sin 10x).
2
1
6. Show that sin 2x cos 3x cos 5x = (sin 4x − sin 6x + sin 10x).
4
Ans: By product-to-sum identities we have
1 1
sin 2x cos 3x cos 5x = sin 2x(cos 2x + cos 8x) = (sin 2x cos 2x + sin 2x cos 8x)
2 2
1
= (sin 4x − sin 6x + sin 10x).
4

1
7. Show that sin 3x sin 4x cos 5x = (− cos 2x + cos 4x + cos 6x − cos 12x).
4
Ans: By product-to-sum identities we have
1 1
sin 3x sin 4x cos 5x = (cos x − cos 7x) cos 5x = (cos x cos 5x − cos 7x cos 5x)
2 2 
1 cos 4x + cos 6x cos 2x + cos 12x
= −
2 2 2
1
= (− cos 2x + cos 4x + cos 6x − cos 12x).
4

cos(x + y) + cos(x − y)
8. Prove that = − cot y.
sin(x − y) − sin(x + y)
cos(x + y) + cos(x − y) 2 cos x cos y cos y
Ans: = =− = − cot y.
sin(x − y) − sin(x + y) −2 cos x sin y sin y
1 cos 2x cot 2y
9. Prove that = + .
tan(x + y) − tan(x − y) 2 sin 2y 2
Ans: Using product-to-sum and sum-to-product identities, we have
1 1
= sin(x+y) sin(x−y)
tan(x + y) − tan(x − y) −
cos(x+y) cos(x−y)
 3

cos(x + y) cos(x − y)
=
sin(x + y) cos(x − y) − cos(x + y) sin(x − y)
1
{cos[(x + y) + (x − y)] + cos[(x + y) − (x − y)]}
= 2
sin[(x + y) − (x − y)]
cos 2x + cos 2y
=
2 sin 2y
cos 2x cot 2y
= +
2 sin 2y 2

x+y sin x + sin y

10. Prove that tan = .
2 cos x + cos y
sin x + sin y 2 sin x+y
2
cos x−y
2
sin x+y
2 x+y
Ans: = x+y x−y = x+y = tan .
cos x + cos y 2 cos 2 cos 2 cos 2 2
x
11. Let t = tan .
2
 x 2t
(a) By considering tan x = tan 2 · , show that tan x = .
2 1 − t2
 x 2 tan x2 2t
Ans: tan x = tan 2 · = 2 x
= .
2 1 − tan 2 1 − t2
(b) By using the result in (a), express sin x and cos x in terms of t.
x x x 2 x 2 tan x2 2t
Ans: sin x = 2 sin cos = 2 tan cos = 2 x
= , and cos x =
2 2 2 2 1 + tan 2 1 + t2
sin x 1 − t2
= .
tan x 1 + t2
(Remark: The result of this question will be useful for integration of trigonometric
function, called t-substitution.)
x 1 + cos x
12. Prove that cot = .
2 sin x
1 + cos x 2 cos2 x2 cos x2 x
Ans: = x x = x = cot .
sin x 2 sin 2 cos 2 sin 2 2
1 + cos x 1 + (1 − tan2 x2 )/(1 + tan2 x2 ) 2 x
Alternative method: = x 2 x
= x = cot .
sin x 2 tan 2 /(1 + tan 2 ) 2 tan 2 2
13. Prove the following identities:
3 − 4 cos 2x + cos 4x
(a) sin4 x = ;
8
Ans: Use double-angle formula twice and we have
2
1 − 2 cos 2x + cos2 2x

4 2 2 1 − cos 2x
sin x = (sin x) = =
2 4
1+cos 4x
1 − 2 cos 2x + 2
=
4
 4

3 − 4 cos 2x + cos 4x
= .
8

10 sin x − 5 sin 3x + sin 5x

(b) sin5 x = ;
16
Ans: Make use of the identity in (a) and thus

16 sin5 x = 2 sin x · 8 sin4 x = 2 sin x(3 − 4 cos 2x + cos 4x)

= 6 sin x − 8 sin x cos 2x + 2 sin x cos 4x
= 6 sin x − 4(sin 3x − sin x) + (sin 5x − sin 3x)
= 10 sin x − 5 sin 3x + sin 5x.

10 sin x − 5 sin 3x + sin 5x

Therefore, sin5 x = .
16
3 + 4 cos 2x + cos 4x
(c) cos4 x = ;
8
Ans: Similar to (a),
2
1 + 2 cos 2x + cos2 2x

4 2 21 + cos 2x
cos x = (cos x) = =
2 4
1+cos 4x
1 + 2 cos 2x + 2
=
4
3 + 4 cos 2x + cos 4x
= .
8

10 cos x + 5 cos 3x + cos 5x

(d) cos5 x = ;
16
Ans: Similar to (b), we use the identity in (c) to derive that

16 cos5 x = 2 cos x · 8 cos4 x = 2 cos x(3 + 4 cos 2x + cos 4x)

= 6 cos x + 8 cos x cos 2x + 2 cos x cos 4x
= 6 cos x + 4(cos 3x + cos x) + (cos 5x + cos 3x)
= 10 cos x + 5 cos 3x + cos 5x.

10 cos x + 5 cos 3x + cos 5x

Therefore, cos5 x = .
16
3 − 4 cos 4x + cos 8x
(e) sin4 x cos4 x = ;
128  4
4 4 4 1 1
Ans: Since sin x cos x = (sin x cos x) = sin 2x = sin4 2x, we re-
2 16
place x by 2x in the identity of (a) to get
1 3 − 4 cos 4x + cos 8x 3 − 4 cos 4x + cos 8x
sin4 x cos4 x = · = .
16 8 128
 5

10 sin 2x − 5 sin 6x + sin 10x

(f) sin5 x cos5 x = .
512
Ans: Using the same technique as in (e), we start from the identity in (b) to
obtain
1 1 10 sin 2x − 5 sin 6x + sin 10x
sin5 x cos5 x = sin5 2x = ·
32 32 16
10 sin 2x − 5 sin 6x + sin 10x
= .
512
1
14. Show that sin2 x cos4 x = (2 + cos 2x − 2 cos 4x − cos 6x).
32
Ans: Decomposition cos4 x into two parts equally and then we can compute that
 2
2 4 2 2 2 1
sin x cos x = sin x cos x · cos x = sin 2x cos2 x
2
1 cos 2x + 1 1 − cos 4x cos 2x + 1
= sin2 2x · = ·
4 2 8 2
1
= (cos 2x + 1 − cos 4x cos 2x − cos 4x)
16  
1 cos 6x + cos 2x
= cos 2x + 1 − − cos 4x
16 2
1
= (2 + cos 2x − 2 cos 4x − cos 6x).
32

15. Prove the following identites (called triple angle formula):

(a) sin 3x = 3 sin x − 4 sin3 x;
Ans:

sin 3x = sin(2x + x)
= sin 2x cos x + cos 2x sin x
= (2 sin x cos x) cos x + (1 − 2 sin2 x) sin x
= 2 sin x(1 − sin2 x) + (1 − 2 sin2 x) sin x
= 3 sin x − 4 sin3 x

(b) cos 3x = 4 cos3 x − 3 cos x;

Ans:

cos 3x = cos(2x + x)
= cos 2x cos x − sin 2x sin x
= (2 cos2 x − 1) cos x − (2 sin x cos x) sin x
= (2 cos2 x − 1) cos x − 2(1 − cos2 x) cos x
= 4 cos3 x − 3 cos x
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3 tan x − tan3 x
(c) tan 3x = .
1 − 3 tan2 x
Ans:

tan 3x = tan(2x + x)
tan 2x + tan x
=
1 − tan 2x
tan x
2 tan x
1−tan2 x
+ tan x
= 2 tan x


1 − 1−tan2 x tan x
 3

3 tan x−tan x
1−tan2 x
=  
1−3 tan2 x
1−tan2 x

3 tan x − tan3 x
=
1 − 3 tan2 x

16. Given that A, B, C, D are four interior angles of a quadrilateral ABCD.

Prove that
A+B A+C A+D
cos A + cos B + cos C + cos D = −4 cos cos cos .
2 2 2
Ans: We have that D = 2π − (A + B + C) and thus

cos A + cos B + cos C + cos D

= cos A + cos B + cos C + cos(A + B + C)
A+B A−B A+B A + B + 2C
= 2 cos cos + 2 cos cos
2  2 2  2
A+B A−B A + B + 2C
= 2 cos cos + cos
2 2 2
A+B A+C B+C
= 4 cos cos cos
2 2  2 
A+B A+C A+D
= 4 cos cos cos π −
2 2 2
A+B A+C A+D
= −4 cos cos cos .
2 2 2

17. If A + B + C = π, show that

(a) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C;

Ans:

sin 2A + sin 2B + sin 2C = 2 sin(A + B) cos(A − B) + sin 2C

= 2 sin(π − C) cos(A − B) + sin 2C
= 2 sin C cos(A − B) + 2 sin C cos C
= 2 sin C[cos(A − B) + cos C]
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A−B+C A−B−C
= 2 sin C · 2 cos cos
2 2
π − 2B 2A − π
= 4 sin C cos cos
2 2
= 4 sin A sin B sin C.

(b) tan A + tan B + tan C = tan A tan B tan C;

Ans:

tan A + tan B + tan C = tan A + tan B + tan(π − A − B)

= tan A + tan B − tan(A + B)
= (1 − tan A tan B) tan(A + B) − tan(A + B)
= − tan A tan B tan(A + B)
= tan A tan B tan C.

(c) cot A cot B + cot B cot C + cot C cot A = 1.

tan A + tan B
Ans: Recall that tan(A + B) = and we have
1 − tan A tan B
cot A cot B + cot B cot C + cot C cot A
1 1  1 1 
= − +
tan A tan B tan(A + B) tan B tan A
1 1 tan A + tan B
= −
tan A tan B tan(A + B) tan A tan B
 
1 tan A + tan B
= 1−
tan A tan B tan(A + B)
= 1.

18. Prove that for any x 6= 2mπ, m is an integer,

sin(n + 21 )x
1 + 2 cos x + 2 cos 2x + 2 cos 3x + · · · + 2 cos nx = .
sin x2

x (2k + 1)x (2k − 1)x

Ans: Note that 2 sin cos kx = sin − sin for all k ≥ 1.
2 2 2
Therefore,
x
sin (1 + 2 cos x + 2 cos 2x + 2 cos 3x + · · · + 2 cos nx)
2
n
x X x
= sin + 2 sin cos kx
2 k=1 2
n  
x X (2k + 1)x (2k − 1)x
= sin + sin − sin
2 k=1 2 2
 8

 
(2n + 1)x 1
= sin = sin n + x.
2 2

sin(n + 12 )x
Hence 1 + 2 cos x + 2 cos 2x + 2 cos 3x + · · · + 2 cos nx = .
sin x2

General Solutions of Trigonometric Equations

19. (General Solutions of Trigonometric Equations)

• If sin x = p, then let α = sin−1 (p), then all solutions of the equation sin x = p
are in form of nπ + (−1)n α where n is an integer;
• If cos x = p, then let α = cos−1 (p), then all solutions of the equation cos x = p
are in form of 2nπ ± α where n is an integer;
• If tan x = p, then let α = tan−1 (p), then all solutions of the equation tan x = p
are in form of nπ + α where n is an integer.

By using the above, solve the following equations.

1
(a) sin x = ;
2
π
Ans: Note that α = satisfies the above equation. Thus,
6
nπ 6n + (−1)n
x = nπ + (−1) = π, where n is an integer.
6 6
√
3
(b) cos x = − ;
2
5π
Ans: Note that α = satisfies the above equation. Thus,
6
5π 12n ± 5
x = 2nπ ± = π, where n is an integer.
6 6
√
(c) tan x = − 3.
2π
Ans: Note that α = satisfies the above equation. Thus,
3
2π 3n + 2
x = nπ + = π, where n is an integer.
3 3
20. Solve the following equations.
1
(a) cos 5x = , where 0 ≤ x < 2π; (Hint: 0 ≤ 5x < 10π.)
2
Ans: From the previous question we have
π
5x = 2nπ ± , where n is an integer. Since 0 ≤ 5x < 10π,
3
π π
5x = 2nπ + for n = 0, 1, 2, 3, 4 or 5x = 2nπ − for n = 1, 2, 3, 4, 5.
3 3
π π 7π 11π 13π 17π 19π 23π 5π 29π
Hence, x = , , , , , , , , , .
15 3 15 15 15 15 15 15 3 15
 9

(b) sin 4x = sin 24◦ , where 0◦ ≤ x < 180◦ ;

2
Ans: 4x = nπ + (−1)n π, where n is an integer.
15
Since 0 ≤ 4x < 4π, n = 0, 1, 2, 3.
π 13π 8π 43π
Hence, x = , , , or expressed in degree as x = 6◦ , 39◦ , 96◦ , 129◦ .
30 60 15 60
(c) tan 3x = 1, where π ≤ x < 2π.
π
Ans: 3x = nπ + ,where n is an integer.
4
Since 3π ≤ 3x < 6π, n = 3, 4, 5.
13π 17π 7π
Hence, x = , , .
12 12 4
21. Solve sin 7x − sin x = cos 4x for 0◦ ≤ x ≤ 180◦ .
Ans: Note that sin 7x − sin x = 2 sin 3x cos 4x and thus
1
2 sin 3x cos 4x = cos 4x =⇒ sin 3x = or cos 4x = 0.
2
π 5π 13π 17π π 3π 5π 7π
By solving the equations, we have x = , , , ; , , , .
18 18 18 18 8 8 8 8
In other words, x = 10◦ , 50◦ , 130◦ , 170◦ ; 22.5◦ , 67.5◦ , 112.5◦ , 157.5◦ .
π
22. Solve sin x sin 2x = cos 3x cos 4x for 0 ≤ x ≤ .
2
Ans: Note that 2 sin x sin 2x = cos x − cos 3x and 2 cos 3x cos 4x = cos 7x + cos x.
Therefore, we have

cos x − cos 3x = cos 7x + cos x

0 = cos 7x + cos 3x
0 = 2 cos 5x cos 2x

Then cos 5x = 0 or cos 2x = 0.

π π 3π π
By solving the equation, we have x = , , , .
10 4 10 2