5 - Q2 Pre Calculus
5 - Q2 Pre Calculus
Pre-Calculus
Quarter 2 - Module 5
Trigonometric Identities
i
About the Module
This module was designed and written with you in mind. It is here to help you master
about Trigonometric Identities. The scope of this module permits it to be used in
many different learning situations. The language used recognizes the diverse
vocabulary level of students. The lessons are arranged based on the Most Essential
Learning Competencies (MELCs) released by the Department of Education (DepEd)
for this school year 2020 – 2021.
ii
What I Know (Pretest)
Instruction: Choose the letter of the correct answers to the following items. Write them
on a separate sheet of paper.
6. What are the identities derived from the sum and difference identities?
A. Double-Angle C. Pythagorean
B. Half-Angle D. Reciprocal
1
10. How do you express 𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 𝜃 in terms of 𝑐𝑜𝑠 𝜃?
A. ±√1 + 𝑐𝑜𝑠 2 𝜃 B. ±√1 − 𝑐𝑜𝑠 2 𝜃 C. 𝑐𝑜𝑠 𝜃 + 1 D. 𝑐𝑜𝑠 𝜃 − 1
24 𝜋
12. Given that sin 𝑡 = 25 and 2
< 𝑡 < 𝜋, what is the exact value 𝑠𝑖𝑛 2𝑡?
225 336
A. − B. − C. 336 D. 225
224 225
24 𝜋
13. Given that sin 𝑡 = 25 and 2
< 𝑡 < 𝜋, what is the exact value 𝑐𝑜𝑠 2𝑡?
565 24 527 7
A. − B. − C. − D.
627 25 625 25
𝜋
14. What is the exact value of 𝑡𝑎𝑛 8
?
A. 2 + √3 B. √2 − 1 C. √3 +1 D.2
𝜃 4
15. What is the exact value of 𝑡𝑎𝑛 ( ) if 𝑡𝑎𝑛 𝜃 = and 𝑐𝑜𝑠 𝜃 < 0 ?
2 3
A. -2 B. 2 C. 0.2 D. √2
2
Lesson Identity and Conditional Equations
1
What’s In
What’s New
Activity 1.1: Find the Impostor!
Instruction: The following crews have expressions equivalent to each other. Your task
is to find the impostor who has an expression not equivalent to the rest of the crews.
After the impostor is identified, write its corresponding name in the box provided
below to reveal a phrase that describes you. Write the phrase you found on your
answer sheet. Good luck!
1 2 3
Cute Not me You Self Me Yup Are
Good
1 2
3(𝑥 + 2) 9𝑥 + 18 2
6(𝑥 + 2𝑥 + 1) 6𝑥 + 12
3𝑥 + 6 2𝑥 + 3 6(𝑥 + 1)2
3 6𝑥 2 + 12𝑥 + 6
3
20𝑥 − 10𝑦 −5(−4𝑥 + 2𝑦)
−5(−4𝑥 − 2𝑦) 5(4𝑥 − 2𝑦)
3
What Is It
3𝑥 − 4 = 8
3𝑥 − 4 + 𝟒 = 8 + 𝟒 Add both sides with positive 4
3𝑥 = 12
3𝑥 12
= Divide both sides with 3
3 3
𝑥=4
1 1 1
b. The equation + = is true only for 𝑝 = −2. Therefore, the equation is
3 𝑝 3𝑝
a conditional equation. To show the solution, we have the following:
1 1 1
+ =
3 𝑝 3𝑝
1 1 1 Multiply both sides with 3p
3𝑝 ( + = ) 3𝑝
3 𝑝 3𝑝
(3𝑝) (3𝑝) 3𝑝
+ =
3 𝑝 3𝑝
𝑝+3=1
𝑝+3−𝟑=1−𝟑 Subtract both sides with 3
𝑝 = −2
c. Note that the left side can be expressed in factored form as (𝑥 + 1)2 .
Similarly, the right side can be expressed as 𝑥 2 + 2𝑥 + 1. Hence, we can
write the equation as follows:
(𝑥 + 1)2 = (𝑥 + 1)2 or 𝑥 2 + 2𝑥 + 1 = 𝑥 2 + 2𝑥 + 1
4
Therefore, the equation is true for all values of 𝑥. Thus, it is an identity
equation.
What’s More
1. 2𝑥 2 = 72
2. (𝑥 + 1)(𝑥 − 1) = 𝑥(𝑥 + 1) − (𝑥 + 1)
1 6
3. + 1 =
𝑛 𝑛
4. (𝑥 + 1)2 −42 = (𝑥 − 3)(𝑥 + 5)
5. sin 𝑥 csc 𝑥 = cos 𝑥 sec 𝑥
5
Lesson Trigonometric Identities
2
What’s In
Recall that the sine and cosine functions (and four others: tangent, cosecant,
secant and cotangent) of angles measuring between 0° and 90° were defined as ratios
of sides of a right angle.
Let 𝜃 be an angle in standard position and 𝑃(𝜃) = 𝑃(𝑥, 𝑦) the point on its
terminal side on the unit circle. Then,
1
sin 𝜃 = 𝑦 csc 𝜃 = 𝑦 , 𝑦 ≠ 0
1
cos 𝜃 = 𝑥 sec 𝜃 = 𝑥 , 𝑥 ≠ 0
𝑦 𝑥
tan 𝜃 = ,𝑥 ≠ 0 cot 𝜃 = ,𝑦 ≠ 0
𝑥 𝑦
Note that all these holds whether we take the values of the variable to
be real numbers (𝑖. 𝑒. , "𝑥") or measures of angles (𝜃).
The sine, cosine and tangent functions are known as the basic circular
functions. The cosecant, secant and cotangent functions are called reciprocal
functions because they are simply the reciprocal of the basic circular functions.
What’s New
√3 1
Let 𝑃 ( , ) be the terminal point of an arc length 𝑠 on the unit circle. Give the
2 2
values of the six circular functions of 𝑠.
sin 𝑠 = csc 𝑠 =
cos 𝑠 = sec 𝑠 =
tan 𝑠 = cot 𝑠 =
6
What Is It
From the definition of the six circular functions of angles measuring between
0° and 90° as defined ratios of the sides of a right angle, the reciprocal and quotient
identities follow. More so, by simplifying a trigonometric expression using the
reciprocal and quotient identities, we obtain the Pythagorean identities.
Reciprocal Identities
1 1 1
𝑠𝑖𝑛 𝜃 = 𝑐𝑠𝑐 𝜃 𝑐𝑜𝑠 𝜃 = 𝑠𝑒𝑐 𝜃 𝑡𝑎𝑛 𝜃 = 𝑐𝑜𝑡 𝜃
Remember!
In addition to the eight identities above, we also have the following identities:
Even-Odd Identities
𝑠𝑖𝑛 (−𝜃) = −𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠 (−𝜃) = 𝑐𝑜𝑠 𝜃 𝑡𝑎𝑛 (−𝜃) = −𝑡𝑎𝑛 𝜃
7
Quotient Identity
Solution:
𝐬𝐢𝐧 𝒙 1 Manipulating
(a) tan 𝑥 csc 𝑥 = ∙
cos 𝑥 𝐬𝐢𝐧 𝒙 Reciprocal Identity
1
=
cos 𝑥
Reciprocal Identity
= sec 𝑥
1 Manipulating
= 𝑠𝑖𝑛2 𝜃 ( )
𝑐𝑜𝑠 2 𝜃 Reciprocal Identity
𝑠𝑖𝑛2 𝜃
=
𝑐𝑜𝑠 2 𝜃
= 𝑡𝑎𝑛2 𝜃 Quotient Identity
1 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙
(c) sin 𝑥 𝑐𝑠𝑐 𝑥 𝑡𝑎𝑛 𝑥 𝑐𝑜𝑡 𝑥 = 𝐬𝐢𝐧 𝒙 ∙ ∙ ∙
𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏 𝒙
Quotient Identities
= 1
sin 𝑥 sin 𝑥
(d) cos (𝑥) = cos 𝑥 cos (𝑥)
cos(−𝑥)
Even-Odd Identity
= sin 𝑥
8
Other Trigonometric Identities
The Fundamental Trigonometric Identities dealt only with one angle. Other
Trigonometric Identities involve two angles, say 𝐴 and 𝐵. The following identities are
used to determine the exact values of the sine and cosine of the sum and difference
of two given angles, 𝛼 and 𝛽.
𝑠𝑖𝑛 (𝛼 + 𝛽) = 𝑠𝑖𝑛 𝛼 𝑐𝑜𝑠 𝛽 + 𝑐𝑜𝑠 𝛼 𝑠𝑖𝑛 𝛽 𝑠𝑖𝑛 (𝛼 − 𝛽) = 𝑠𝑖𝑛 𝛼 𝑐𝑜𝑠 𝛽 − 𝑐𝑜𝑠 𝛼 𝑠𝑖𝑛 𝛽
𝑐𝑜𝑠 (𝛼 + 𝛽) = 𝑐𝑜𝑠 𝛼 𝑐𝑜𝑠 𝛽 − 𝑠𝑖𝑛 𝛼 𝑠𝑖𝑛 𝛽 𝑐𝑜𝑠 (𝛼 − 𝛽) = 𝑐𝑜𝑠 𝛼 𝑐𝑜𝑠 𝛽 + 𝑠𝑖𝑛 𝛼 𝑠𝑖𝑛 𝛽
𝑡𝑎𝑛 𝛼 + 𝑡𝑎𝑛 𝛽 𝑡𝑎𝑛 𝛼 − 𝑡𝑎𝑛 𝛽
𝑡𝑎𝑛 (𝛼 + 𝛽) = 𝑡𝑎𝑛 (𝛼 − 𝛽) =
1 − 𝑡𝑎𝑛 𝛼 𝑡𝑎𝑛 𝛽 1 + 𝑡𝑎𝑛 𝛼 𝑡𝑎𝑛 𝛽
Example 2.2:
Find the exact value of the following using Sum and Difference Identities.
(a) 𝑠𝑖𝑛 15°
𝜋
(b) 𝑐𝑜𝑠 12
3 12
(c) 𝑡𝑎𝑛 (𝛼 + 𝛽) if 𝑠𝑖𝑛 𝛼 = and 𝑐𝑜𝑠 𝛽 = − , and 𝛼 is in Quadrant I while 𝛽 is in
5 13
Quadrant II.
Solution:
(a) The given angle can be expressed in terms of two special angles. That is,
15° = 45° − 30°.
Use the formula for the sine of the difference of the angles to obtain the
exact value of 𝑠𝑖𝑛 15°.
Difference
𝑠𝑖𝑛 15° = 𝑠𝑖𝑛 (45° − 30°)
Identity for Sine
= sin 45°𝑐𝑜𝑠 30° − 𝑐𝑜𝑠 45° 𝑠𝑖𝑛 30°
Apply unit
triangle
√2 √3 √2 1 property or
=( )( ) − ( )( )
2 2 2 2 use a
scientific
√6 √2
= − calculator.
4 4
√6 − √2
=
4
9
𝜋 𝜋 𝜋 𝜋 𝜋 𝜋
(b) Observe that = − . Thus, 𝑐𝑜𝑠 = 𝑐𝑜𝑠 ( − ).
12 3 4 12 3 4
𝜋 𝜋 𝜋 Difference Identity
𝑐𝑜𝑠 = 𝑐𝑜𝑠 ( − )
12 3 4 for Cosine
𝜋 𝜋 𝜋 𝜋
= 𝑐𝑜𝑠 𝑐𝑜𝑠 + 𝑠𝑖𝑛 𝑠𝑖𝑛
3 4 3 4 Apply unit
triangle
1 √2 √3 √2 property or
= ( )( ) + ( )( )
2 2 2 2 use a
√2 √6 scientific
= + calculator.
4 4
√2 + √6
=
4
(c) To find 𝑡𝑎𝑛 𝛼, use the Pythagorean Identity 𝑠𝑖𝑛2 𝛼 + 𝑐𝑜𝑠 2 𝛼 = 1 and solve for
the value of 𝑐𝑜𝑠 𝛼.
𝑠𝑖𝑛2 𝛼 + 𝑐𝑜𝑠 2 𝛼 = 1
3 2
Given ( ) + 𝑐𝑜𝑠 2 𝛼 = 1
5
9
+ 𝑐𝑜𝑠 2 𝛼 = 1
25
9
𝑐𝑜𝑠 2 𝛼 = 1 −
25
25 − 9
𝑐𝑜𝑠 2 𝛼 =
25
16
𝑐𝑜𝑠 2 𝛼 =
25
16
√𝑐𝑜𝑠 2 𝛼 = √
25
4
𝑐𝑜𝑠 𝛼 = ±
5
𝟒
From the given, 𝛼 is in quadrant I, thus 𝒄𝒐𝒔 𝜶 = 𝟓. To find the value of 𝑡𝑎𝑛 𝛼,
𝑠𝑖𝑛 𝛼
use the equation 𝑡𝑎𝑛 𝛼 = 𝑐𝑜𝑠 𝛼.
3
5
𝑡𝑎𝑛 𝛼 = 4
5
3 𝟓
=∙
𝟓 4
𝟑
𝒕𝒂𝒏 𝜶 =
𝟒
10
Do the same procedure as presented above to find the value of 𝑡𝑎𝑛 𝛽.
To find 𝑡𝑎𝑛 𝛽, use the Pythagorean Identity 𝑠𝑖𝑛2 𝛽 + 𝑐𝑜𝑠 2 𝛽 = 1 and solve
for the value of 𝑠𝑖𝑛 𝛽.
𝑠𝑖𝑛2 𝛽 + 𝑐𝑜𝑠 2 𝛽 = 1
2
12 2
𝑠𝑖𝑛 𝛽+ (− ) = 1
13
Given
144
𝑠𝑖𝑛2 𝛽 + =1
169
144
𝑠𝑖𝑛2 𝛽 = 1 −
169
169 − 144
𝑠𝑖𝑛2 𝛽 =
169
25
𝑠𝑖𝑛2 𝛽 =
169
25
√ 𝑠𝑖𝑛2 𝛽 = √
169
5
𝑠𝑖𝑛 𝛽 = ±
13
𝟓
From the given, 𝛽 is in quadrant II, thus 𝒔𝒊𝒏 𝛽 = . To find the value of 𝑡𝑎𝑛 𝛽,
𝟏𝟑
𝑠𝑖𝑛 𝛽
use the equation 𝑡𝑎𝑛 𝛽 = .
𝑐𝑜𝑠 𝛽
5
13
𝑡𝑎𝑛 𝛽 = 12
(− )
13
5 𝟏𝟑
= ∙ (− )
𝟏𝟑 12
𝟓
𝒕𝒂𝒏 𝜷 = −
𝟏𝟐
𝑡𝑎𝑛 𝛼 + 𝑡𝑎𝑛 𝛽
𝑡𝑎𝑛 (𝛼 + 𝛽) =
1 − 𝑡𝑎𝑛 𝛼 𝑡𝑎𝑛 𝛽
3 + (− 5 )
=
4 12
1 − ( ) (− 5 )
3
4 12
3− 5
= 4 12
1 + 15
48
11
9−5 You can
= 12 reduce the
48 + 15 numerator
48 and
4 denominat
= 12 or by
63
48 removing
4 𝟒𝟖 its
= ∙ common
𝟏𝟐 63
value.
16
𝑡𝑎𝑛 (𝛼 + 𝛽) =
63
Double-Angle Identities
𝑐𝑜𝑠 2𝛼 = 2𝑐𝑜𝑠 2 𝛼 − 1
𝑐𝑜𝑠 2𝛼 = 1 − 𝑠𝑖𝑛2 𝛼
Example 2.3:
4 𝜋
If 𝑠𝑖𝑛 𝛼 = 5 and 2
< 𝛼 < 𝜋, find the exact values of the following:
(a) 𝑠𝑖𝑛 2𝛼
(b) 𝑐𝑜𝑠 2𝛼
(c) 𝑡𝑎𝑛 2𝛼
Solution:
𝜋 𝟑
Because 2
< 𝛼 < 𝜋, then 𝒄𝒐𝒔 𝜶 = − 𝟓. Therefore,
4
5
𝑡𝑎𝑛 𝛼 = 3
−
5
4 𝟓
= ∙ (− )
𝟓 3
𝟒
=−
𝟑
Then,
(a) 𝑠𝑖𝑛 2𝛼 = 2𝑠𝑖𝑛 𝛼 𝑐𝑜𝑠 𝛼
4 3
𝑠𝑖𝑛 2𝛼 = 2 ( ) (− )
5 5
24
=−
25
3 2 4 2
𝑐𝑜𝑠 2 α = (− ) − ( )
5 5
9 16
= −
25 25
7
=−
25
2𝑡𝑎𝑛 𝛼
(c) 𝑡𝑎𝑛 2𝛼 =
1−𝑡𝑎𝑛2 𝛼
4
2(− )
3
𝑡𝑎𝑛 2𝛼 =
4 2
1−(−3)
8
−3
= 16
1− 9
8
−3
= 9−16
9
8
−3
= −7
9
8 9
= − ∙−
3 7
24
=
7
13
𝛼
The half-angle identities can be derived by replacing 𝛼 with in the double-
2
angle identity for cosine.
Half-Angle Identities
𝛼 1 − 𝑐𝑜𝑠 𝛼 𝛼 1−𝑐𝑜𝑠 𝛼
𝑠𝑖𝑛 = ±√ 𝑡𝑎𝑛 2
= ±√ ; 𝑐𝑜𝑠 𝛼 ≠ 1
2 2 1+𝑐𝑜𝑠 𝛼
𝛼 1 + 𝑐𝑜𝑠 𝛼 𝛼 1−𝑐𝑜𝑠 𝛼
𝑐𝑜𝑠
2
= ±√
2
𝑡𝑎𝑛
2
= ; 𝑠𝑖𝑛 𝛼 ≠ 0
𝑠𝑖𝑛 𝛼
𝛼 𝑠𝑖𝑛 𝛼
𝑡𝑎𝑛
2
= ; 𝑐𝑜𝑠 𝛼 ≠ −1
1−𝑐𝑜𝑠 𝛼
Example 2.4:
Solution:
𝛼 5𝜋 5𝜋
(a) Let 2
= 8
. Then 8
is in quadrant II where the sine is positive.
5𝜋 5𝜋 𝛼 1−𝑐𝑜𝑠 𝛼
Also, 𝛼 = 2 ( 8 ) = 4 . Since sine is positive, 𝑠𝑖𝑛 2 = √ 2 .
5𝜋
5𝜋 √1 − 𝑐𝑜𝑠 4
𝑠𝑖𝑛 =
8 2
√2
1 − (− )
√ 2
=
2
√2
√1 + 2
=
2
2 + √2
√ 2
=
2
2 + √2 1
=√ ∙
2 2
5𝜋 2 + √2
𝑠𝑖𝑛 =√
8 4
14
𝜃
(b) Since 𝜃 is in quadrant I, 𝑐𝑜𝑠 is positive.
2
𝜃 1 + 𝑐𝑜𝑠 𝛼
𝑐𝑜𝑠 =√
2 2
3
𝜃 √1 + 5
𝑐𝑜𝑠 =
2 2
5+3
√ 5
=
2
8
√5
=
2
8 1
=√ ∙
5 2
4
=√
5
𝜃 2√5
𝑐𝑜𝑠 =
2 5
𝜃
(c) Note that 180° < 𝜃 < 270°, then 90° < < 135° which is in quadrant II.
2
𝜃
Therefore, 𝑡𝑎𝑛 2
must be negative.
First, solve the values of 𝑠𝑖𝑛 𝜃 and 𝑐𝑜𝑠 𝜃. Tha value of 𝑐𝑜𝑠 𝜃 can
be solved using the formula 1 + 𝑡𝑎𝑛2 𝜃 = 𝑠𝑒𝑐 2 𝜃.
1 + (2)2 = 𝑠𝑒𝑐 2 𝜃
1 + 4 = 𝑠𝑒𝑐 2 𝜃
𝑠𝑒𝑐 2 𝜃 = 5
√𝑠𝑒𝑐 2 𝜃 = √5
𝑠𝑒𝑐 𝜃 = ±√5
𝜃
Solving for the value of 𝑡𝑎𝑛 2 ,
𝜃 𝑠𝑖𝑛 𝜃
𝑡𝑎𝑛 =
2 1 − 𝑐𝑜𝑠 𝜃
15
2√5
−
= 5
√5
1 − (− )
5
2√5
−
= 5
√5
1+
5
2√5
−
= 5
5 + √5
5
2√5 𝟓
=− ∙
𝟓 5 + √5
2√5
=−
5 + √5
What’s More
𝛼 12
3. 𝑐𝑜𝑠 2 if 𝑐𝑜𝑠 𝛼 = − 13 and 180° < 𝛼 < 270°
16
What I Need to Remember
• An identity is an equation that is true for all real numbers and whose two sides
are both defined. A conditional equation, on the other hand, is an equation that
is true only for at least one real number.
• When simplifying equations that involve trigonometric expressions, it is
necessary to consider the following fundamental trigonometric identities:
A. Reciprocal Identities
1 1 1
𝑠𝑖𝑛 𝜃 = 𝑐𝑠𝑐 𝜃 𝑐𝑜𝑠 𝜃 = 𝑠𝑒𝑐 𝜃 𝑡𝑎𝑛 𝜃 = 𝑐𝑜𝑡 𝜃
B. Quotient Identities
𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠 𝜃
𝑡𝑎𝑛 𝜃 = 𝑐𝑜𝑠 𝜃 𝑐𝑜𝑡 𝜃 = 𝑠𝑖𝑛 𝜃
C. Pythagorean Identities
𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 𝜃 = 1 1 + 𝑡𝑎𝑛2 𝜃 = 𝑠𝑒𝑐 2 𝜃 1 + 𝑐𝑜𝑡 2 𝜃 = 𝑐𝑠𝑐 2 𝜃
D. Even-Odd Identities
s𝑖𝑛 (−𝜃) = −𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠 (−𝜃) = 𝑐𝑜𝑠 𝜃 𝑡𝑎𝑛 (−𝜃) = −𝑡𝑎𝑛 𝜃
• Other trigonometric identities involve functions of the sum or difference of
two angles, double angle, or half angle. The following are the sum and
difference identities for sine, cosine and tangent functions.
𝑠𝑖𝑛 (𝛼 + 𝛽) = 𝑠𝑖𝑛 𝛼 𝑐𝑜𝑠 𝛽 + 𝑐𝑜𝑠 𝛼 𝑠𝑖𝑛 𝛽
𝑠𝑖𝑛 (𝛼 − 𝛽) = 𝑠𝑖𝑛 𝛼 𝑐𝑜𝑠 𝛽 − 𝑐𝑜𝑠 𝛼 𝑠𝑖𝑛 𝛽
𝑐𝑜𝑠 (𝛼 − 𝛽) = 𝑐𝑜𝑠 𝛼 𝑐𝑜𝑠 𝛽 + 𝑠𝑖𝑛 𝛼 𝑠𝑖𝑛 𝛽 𝑐𝑜𝑠 (𝛼 + 𝛽) = 𝑐𝑜𝑠 𝛼 𝑐𝑜𝑠 𝛽 − 𝑠𝑖𝑛 𝛼 𝑠𝑖𝑛 𝛽
• From these identities, the double-angle identities which are shown below,
can be derived.
𝑠𝑖𝑛 2𝛼 = 2𝑠𝑖𝑛 𝛼 𝑐𝑜𝑠 𝛼 2𝑡𝑎𝑛 𝛼
𝑡𝑎𝑛 2𝛼 =
1−𝑡𝑎𝑛2 𝛼
𝑐𝑜𝑠 2𝛼 = 𝑐𝑜𝑠 2 𝛼 − 𝑠𝑖𝑛2 𝛼
𝑐𝑜𝑠 2𝛼 = 2𝑐𝑜𝑠 2 𝛼 − 1
𝑐𝑜𝑠 2𝛼 = 1 − 𝑠𝑖𝑛2 𝛼
𝛼 1 − 𝑐𝑜𝑠 𝛼 𝛼 1−𝑐𝑜𝑠 𝛼
𝑠𝑖𝑛 = ±√ 𝑡𝑎𝑛
2
= ±√ ; 𝑐𝑜𝑠 𝛼 ≠ 1
2 2 1+𝑐𝑜𝑠 𝛼
𝛼 1 + 𝑐𝑜𝑠 𝛼 𝛼 1−𝑐𝑜𝑠 𝛼
𝑐𝑜𝑠
2
= ±√
2
𝑡𝑎𝑛 2
= ; 𝑠𝑖𝑛 𝛼 ≠ 0
𝑠𝑖𝑛 𝛼
𝛼 𝑠𝑖𝑛 𝛼
𝑡𝑎𝑛 2
= ; 𝑐𝑜𝑠 𝛼 ≠ −1
1−𝑐𝑜 𝑠 𝛼
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What I Can Do
𝑥2 − 4 13 1 1 11
𝑥2 − 1 = 0 = 2𝑥 − 1 𝑥+ = 𝑥−
𝑥−2 6 3 2 2
(𝑥 + 7)2 = 𝑥 2 + 14𝑥 + 49 (𝑥 + 7)2 = 𝑥 2 + 49 1 1 1
( 𝑥 − √3) ( 𝑥 + √3) = 𝑥 2 − 3
𝑥 2 − 1 = (𝑥 − 1)(𝑥 + 1) 2 2 4
𝑥2 − 4
=𝑥+2
𝑥−2
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Assessment (Posttest)
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10. How do you express 𝑠𝑒𝑐 𝜃 in terms of 𝑐𝑜𝑠 𝜃?
1
𝐴. 𝑐𝑜𝑠 𝜃
B. 𝑐𝑜𝑠 𝜃 − 1 C. 𝑐𝑜𝑠 𝜃 + 1 D. 𝑐𝑜𝑠 𝜃
3 𝜋
12. Given that sin 𝑡 = 5 and 2
< 𝑡 < 𝜋, what is the exact value 𝑠𝑖𝑛 2𝑡?
25 24 7
𝐴. − 24 B. − 25 C. 25 D. 25
3 𝜋
13. Given that sin 𝑡 = 5 and 2
< 𝑡 < 𝜋, what is the exact value 𝑐𝑜𝑠 2𝑡?
25 24 7
𝐴. − 24 B. − 25 C. 25 D. 25
14. What is the exact value of 𝑡𝑎𝑛 75° using the Half-Angle Identity?
𝐴. 2 + √3 B. 2 − √3 C. √3 D.2
4 𝜃
15. If 𝑡𝑎𝑛 𝜃 = and 𝑐𝑜𝑠 𝜃 < 0. What is the exact value of 𝑡𝑎𝑛 ( ) ?
3 2
A. 2 B. -2 C. 0.2 D. √2
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References
Book
Cabral, E.A. et.al, Application of Trigonometric Functions in Periodic Phenomena,
Precalculus (Ateneo de Manila Press,2010), 270-291
Albay, E. M.,Senior High Schoo Series: Pre Calculus (DIWA Learning System, Inc.,
2016), 169-194
PDF File
Garces, I. J.,Mathematical Induction, Precalculus: Learner’s Material for Senior High
School (Quezon City © 2016), 161-170
Images
Impostor Crew Images, retrieved from tinyurl.com/3d733wx5 on January 27, 2021
Congratulations!
You are now ready for the next module. Always remember the following:
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