SodaPDF-converted-My-Report-on-Trigonometry.pptx

UNIT 5:
TRIGONOMETRIC
IDENTITIES

LEARNING OBJECTIVES
1. Derive the identities for cosine, sine, and tangent of a sum
or difference of two real numbers or angles;
2. Derive double and half number identities;
3. Derive sum and product identities;
4. Solve problems involving identities;
5. Solve problems in reduction identities; and
6. Prove a given identity.
ľíigo⭲omctíic Idc⭲titics

IDENTITIES INVOLVING COSINE, SINE, AND TANGENT OF A
SUM OR DIFFERENCE OF TWO REAL NUMBERS
The first identity we shall establish involves cos(𝛼 − 𝛽)
which is the basis of the rest of the identities discussed in this
unit.
Given two points P1(𝑥1, 𝑦1) and P2(𝑥2, 𝑦2) in the rectangular
coordinate system, the distance between these points is
given by
2 +
𝑃1𝑃2 = 𝑥1 − 𝑥2 𝑦1 − 𝑦2
2
𝑃1𝑃2 = 𝑥2 − 𝑥1
2 +
or
𝑦2 − 𝑦1
2

Consider the unit circle and the arcs of the unit circle of
lengths 𝛼 and 𝛽 starting from point 𝐴(1,0) as shown in
Figure 5.1a. The terminal point determined by the arc length
𝛼 will be
P1 𝛼 = 𝑥1, 𝑦1 = (𝑐𝑜𝑠𝛼, sin𝛼)
And the terminal point determined by 𝛽 will be
P2 𝛽 = 𝑥2, 𝑦2 = (𝑐𝑜𝑠𝛽, sin𝛽)

The distance between the terminal points can be obtained
using the distance formula where the coordinate of the
points are expressed in terms of the real numbers 𝛼 and 𝛽 .
Hence, we have

If we subtract the arc length 𝛼 from the arc length 𝛽, we get the
arc length (𝛽- 𝛼).
Next, we lay an arc starting from the point
𝐴(1,0) whose length is (𝛽- 𝛼) as shown in
Figure 5.2. The endpoint of this arc will
therefore have coordinates
[cos(𝛽- 𝛼), sin(𝛽- 𝛼)].

The length of the chord 𝑃𝐴 determined by the arc length (𝛽- 𝛼)
will be
Since equal arcs determine equal chords, we have 𝑃𝐴 = 𝑃1𝑃2 . It
follows that 2 − 2𝑐𝑜𝑠(𝛼 − 𝛽) = 2 − 2 cos 𝛼 cos 𝛽 − 2 sin 𝛼 sin 𝛽 .

Squaring both sides of the previous equation, we have
2 − 2𝑐𝑜𝑠 𝛼 − 𝛽
−2𝑐𝑜𝑠 𝛼 − 𝛽
−2𝑐𝑜𝑠 𝛼 − 𝛽
𝑐𝑜𝑠 𝛼 − 𝛽
= 2 − 2 cos 𝛼 cos 𝛽 − 2 sin 𝛼 sin 𝛽
= −2 cos 𝛼 cos 𝛽 − 2 sin 𝛼 sin 𝛽
= −2(cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽)
= (cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽)
Hence, the identity for 𝑐𝑜𝑠 𝛼 − 𝛽 is given by
𝑐𝑜𝑠 𝛼 − 𝛽 = (cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽)

To establish the identity for 𝑐𝑜𝑠 𝛼 + 𝛽 , we note that
𝑐𝑜𝑠 𝛼 + 𝛽 = cos[𝛼 − (−𝛽)]
= cos 𝛼 cos(−𝛽) + sin 𝛼 sin(−𝛽)
Note that, cos −𝛽 = cos 𝛽 and sin(−𝛽) = − sin 𝛽, thus
= cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽
Hence, the identity for 𝑐𝑜𝑠 𝛼 + 𝛽 is given by
𝑐𝑜𝑠 𝛼 + 𝛽 = cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽

To derive the identity for sin 𝛼 − 𝛽 , we note that
cos(
𝜋
2
− 𝛼) = cos
𝜋
2
cos 𝛼 + sin
𝜋
2
sin 𝛼
Hence, cos[
𝜋
2
sin 𝛼 − 𝛽 = cos[
= 0(cos 𝛼) +1(sin 𝛼)
= sin 𝛼
− ( 𝛼 − 𝛽)] =sin 𝛼 − 𝛽 or
𝜋
2
− ( 𝛼 − 𝛽)]

Expanding the right hand side we have,
𝜋
2
sin 𝛼 − 𝛽 = cos[ − ( 𝛼 − 𝛽)]
𝜋
2
− 𝛼) + 𝛽]
= cos(
= cos[(
𝜋
2
− 𝛼) ⋅ cos 𝛽 − sin(
𝜋
2
− 𝛼) ⋅ sin 𝛽
= sin 𝛼 cos 𝛽 − cos 𝛼 sin 𝛽
Therefore, we have the following identity for sin 𝛼 − 𝛽 :
sin 𝛼 − 𝛽 = sin 𝛼 cos 𝛽 − cos 𝛼 sin 𝛽

as follows,
Finally, we establish the identity for sin 𝛼 + 𝛽
sin 𝛼 + 𝛽 = sin[𝛼 − (−𝛽)]
= sin 𝛼 cos(−𝛽) − cos 𝛼 sin(−𝛽)
= sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽
(Since cos −𝛽 = cos 𝛽 and sin −𝛽 = −sin 𝛽
Therefore, we have the following identity for sin 𝛼 + 𝛽 :
sin 𝛼 + 𝛽 = sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽

For the identity of tangent of sum of two real numbers, we have
tan 𝛼 + 𝛽 = sin 𝛼+𝛽
= sin 𝛼 cos 𝛽+ cos 𝛼 sin 𝛽
𝑐𝑜𝑠 𝛼+𝛽 cos 𝛼 cos 𝛽− sin 𝛼 sin 𝛽
=
sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽
cos 𝛼 cos 𝛽
cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽
cos 𝛼 cos 𝛽
=
tan 𝛼 + tan 𝛽
1 − tan 𝛼 tann 𝛽

For the identity of tangent of difference of two real numbers, we
have
tan 𝛼 − 𝛽 = sin 𝛼−𝛽
= sin 𝛼 cos 𝛽− cos 𝛼 sin 𝛽
𝑐𝑜𝑠 𝛼−𝛽 cos 𝛼 cos 𝛽+ sin 𝛼 sin 𝛽
=
sin 𝛼 cos 𝛽 − cos 𝛼 sin 𝛽
cos 𝛼 cos 𝛽
cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽
cos 𝛼 cos 𝛽
tan 𝛼 − tan 𝛽
=
1 + tan 𝛼 tann 𝛽

EXAMPLES
1. Find the exact value of cos
Solution:
𝜋
12
.
Rewrite the given angle as a sum or difference of two special
12 12 12
𝜋 4𝜋 3𝜋 𝜋 𝜋
3 4
= − = − .
angles, that is,
Thus,
𝜋
12
cos = cos
𝜋
− 𝜋
3 4 3 4 3 4
= (cos 𝜋
) (cos 𝜋
) + (𝑠𝑖𝑛 𝜋
) (𝑠𝑖𝑛 𝜋
)
=
1
2
2
2
+
3 2
=
2 2 4
2+ 6

EXAMPLES
2. Find the exact value of sin
Solution:
5𝜋
12
.
Rewrite the given angle as a sum of or difference of two
12 12 12
5𝜋 2𝜋 3𝜋 𝜋 𝜋
6 4
= + = + .
special angles, that is,
Thus,
5𝜋
12
sin = sin
𝜋
+ 𝜋
6 4 6 4 6
= (𝑠𝑖𝑛 𝜋
) (cos 𝜋
) + (𝑐𝑜𝑠 𝜋
)(𝑠𝑖𝑛 𝜋
)
=
1
2
2
2
+
3
=
4
2 2+ 6
2 2 4

EXAMPLES
3. Find the exact value of sin(15°).
Solution:
special angles, that is, 15° = 45° − 30°.
Thus,
sin 15° = sin 45° − 30° = (𝑠𝑖𝑛 45°) (cos 30°) − (𝑐𝑜𝑠 45°) (𝑠𝑖𝑛 30°)
=
2 3
−
2
2 1
2
=
2 2
6− 2
4

EXAMPLES
4. Find the exact value of sin 75° .
Solution:
special angles, that is, 75° = 45° + 30°.
Thus,
sin 75° = sin 45° + 30° = (𝑠𝑖𝑛 45°) (cos 30°) + (𝑐𝑜𝑠 45°) (𝑠𝑖𝑛 30°)
=
2 3
+
2
2 1
2
=
2 2
6+ 2
4

EXAMPLES
5. Find the exact value of tan(105°).
Solution:
special angles, that is, 105° = 45° + 60°.
Thus,
tan 105° = tan 45° + 60° =
tan 45° +tan 60°
1−tan 45° tan 60°
=
1+ 3
1− 3
=
1+ 3 1+
1− 3 1+ 3
3
= 1+2 3+3
= 4+2
1−3 −2
3
= −2 − 3

EXAMPLES
+
6. Find the exact value of tan
Solution:
𝜋
12
.
Rewrite the given angle as a sum of or difference of two special
12 12 12
𝜋 4𝜋 3𝜋 𝜋 𝜋
3 4
= − = − .
angles, that is,
Thus,
tan
𝜋
12
= tan −
𝜋 𝜋
3 4
tan −tan
𝜋
𝜋
1+tan tan 𝜋
4
1
1+
3
1
−1 1− 3
3+1
3
= 3 4
= 3
= 3
= 1− 3
1+ 3
=
𝜋
3
1− 3 1− 3
= 1−2 3+3
= 4−2
1+ 3 1− 3 1−3 −2
3
= −2 + 3 = 3 − 2

EXAMPLES
7. Simplify the following expressions using the appropriate identity.

EXAMPLES
Solutions:
For a, we use this identity: 𝑐𝑜𝑠 𝛼 − 𝛽 = (cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽)
cos 8𝑥 cos 𝑥 + sin 8𝑥 sin 𝑥 = cos 8𝑥 − 𝑥 = cos 7𝑥
For b, we use this identity: sin 𝛼 + 𝛽 = sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽
sin 5𝑥 cos 2𝑥 + cos 5𝑥 sin 2𝑥 = sin 5𝑥 + 2𝑥 = sin 7𝑥
For c, we use this identity: 𝑐𝑜𝑠 𝛼 + 𝛽 = (cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽)
sin 4𝑥 sin 6𝑥 − cos 4𝑥 cos 6𝑥 = −(cos 4𝑥 cos 6𝑥 − sin 4𝑥 sin 6𝑥)
= −cos 4𝑥 + 6𝑥 = −𝑐𝑜𝑠 (10𝑥)

EXAMPLES
Solutions:
For d, we use this identity:sin 𝛼 − 𝛽 = sin 𝛼 cos 𝛽 − cos 𝛼 sin 𝛽
sin 3𝑥 cos 4𝑥 − cos 3𝑥 sin 4𝑥 = sin 3𝑥 − 4𝑥 = sin −𝑥 = − sin 𝑥
For e, we use this identity:tan 𝛼 + 𝛽 =
tan 𝛼+tan 𝛽
1−tan 𝛼 tan 𝛽
tan 8𝑥+tan 2x = tan 8𝑥 + 2𝑥 = tan(10𝑥)
1−tan 8𝑥 tan 2x
For f, we use this identity:tan 𝛼 − 𝛽 =
tan 𝛼−tan 𝛽
1+tan 𝛼 tan 𝛽
1+tan 4𝑥 tan 5x
tan 4𝑥−tan 5x = tan 4𝑥 − 5𝑥
= tan(−𝑥) = − tan(𝑥)

EXAMPLES
8. (a) Prove that sin
Proof:
𝜋
2
− 𝑥 = cos 𝑥 and (b) Prove that tan
3𝜋
2
+ 𝑥 = −co𝑡 𝑥
(a) sin
𝜋
2
− 𝑥 = sin
𝜋
2
cos 𝑥 + cos
𝜋
2
sin 𝑥 = cos 𝑥
(b)
= − cos 𝑥+0
= − cot 𝑥
0+sin 𝑥

EXAMPLES
9. If tan 𝛼 =
4
3
, 𝛼 terminates in QIII and tan 𝛽 =
4
3
, 𝛽 terminates in QII, find
tan 𝛼 + 𝛽 and the quadrant where 𝛼 + 𝛽 terminates.
Solution:

EXAMPLES
which means that 𝛼 + 𝛽 is either in QI or QIV. Since 𝛼 + 𝛽 =
1
2
, a
positive real number, it follows that 𝛼 + 𝛽 terminates in QI.

REDUCTION IDENTITIES
Now, we will consider some identities which will help us find
values of trigonometric functions of angles greater than 90° or greater
than 360°.
1. cos (90° − 𝑥) = cos 90° cos 𝑥 + sin 90° sin 𝑥 = sin 𝑥 = cos (
𝜋
2
− 𝑥)
𝜋
2
2. sin (90° − 𝑥) = sin 90° cos 𝑥 − cos 90° sin 𝑥 = cos 𝑥 = sin ( − 𝑥)
cos (90°−𝑥)
𝜋
sin 𝑥 2
sin (90°−𝑥) cos 𝑥
3. tan (90° − 𝑥) = = = cot 𝑥 = tan ( − 𝑥)
These are called reduction identities.

REDUCTION IDENTITIES

REDUCTION IDENTITIES Examples
Using only the reduction identities, find the exact values of the
trigonometric function of the given angle:
(a) cos 600°
Solution:
cos 600 ° = cos(630° − 30°) = cos(7(90°) − 30°)
= ± sin 30 °
= − sin 30 °
= − 1
2

REDUCTION IDENTITIES Examples
Using only the reduction identities, find the exact values of the
trigonometric function of the given angle:
(b) sin 330 °
Solution:
sin 330 ° = sin(360° − 30°) = cos(4(90°) − 30°)
= ± sin 30 °
= − sin 30 °
= − 1
2

DOUBLE ANGLE IDENTITIES
The sum and difference identities can be used to derive the
double angle identities. Consider the expression cos 2𝜃. Note that
cos 2𝜃 = cos 𝜃 + 𝜃 . Applying the sum identity, we have
cos 2𝜃 = cos (𝜃 + 𝜃) = cos 𝜃 cos 𝜃 − sin 𝜃 sin 𝜃 = cos2 𝜃 − sin2 𝜃.
Since sin2 𝜃 = 1 − cos2 𝜃 , we can write
cos 2𝜃 = cos2 𝜃 − (1 − cos2 𝜃) = 2cos2 𝜃 − 1.
Also, since cos2 𝜃 = 1 − sin2 𝜃 , we can write
cos 2𝜃 = 1 − sin2 𝜃 − sin2 𝜃 = 1 − 2 sin2 𝜃 .

DOUBLE ANGLE IDENTITIES
Similarly, sin 2𝜃 = sin 𝜃 + 𝜃 . Applying the sum identity, we
have
sin 2𝜃 = sin (𝜃 + 𝜃) = sin 𝜃 cos 𝜃 + cos 𝜃 sin 𝜃 = 2 sin 𝜃 cos 𝜃.
For the tangent function, we have
tan 2𝜃 = tan (𝜃 + 𝜃) =
tan 𝜃+tan 𝜃
1−tan 𝜃tan 𝜃
=
2tan 𝜃
1−tan2𝜃
.

DOUBLE ANGLE IDENTITIES Examples
Find the values of sin 2t, cos 2t and tan 2t if
(a) sin t =
4
5
and 0 < 𝑡 <
1
2
𝜋

(b) cos t = −
5
13
and
1
2
𝜋 < 𝑡 < 𝜋

(b) tan t =
8
15
and sin 𝑡 < 0

HALF ANGLE IDENTITIES
The double angle identities cos 2𝜃 = 2cos2 𝜃 − 1 and cos 2𝜃 =
1 − 2 sin2 𝜃 can be manipulated to give rise to the half angle
identities. Solving for cos 𝜃 and sin 𝜃 from these equations will yield
the equations
cos2 𝜃 = 1+cos 2𝜃
sin2 𝜃 = 1−cos 2𝜃
2 2
and .
Hence, 𝑐𝑜𝑠 𝜃 = ±
1+cos 2𝜃
2
and 𝑠𝑖𝑛 𝜃 = ±
1−cos 2𝜃
2

Note that for these equations, the angle on the left-hand side is
half of the angle of the right-hand side. Hence, if we let 𝛼 = 2𝜃, then
𝜃 = 𝛼
. Using this value of 𝜃, we have the following half-angle or half-
2
number identities:
𝑐𝑜𝑠
𝛼
2
= ±
1+cos𝛼
2
, 𝑠𝑖𝑛
𝛼
2
= ±
1−cos𝛼
2
and 𝑡𝑎𝑛
𝛼
2
= ±
1−cos𝛼
1+cos𝛼

Also, for tan
𝛼
2
, we can write it in another form as follows:
or

HALF ANGLE IDENTITIES Examples
Use the half-measure identities to find the exact value of the
following:
(a) sin (b) cos (c) tan
𝜋 7𝜋 𝜋
8 8 12
(d) cos 15 ° (e) tan 22.5 °

PRODUCT IDENTITIES
The product identities can be derived using the identities
involving the sine and cosine of a sum or difference of two numbers.

PRODUCT IDENTITIES
The sum and product identities can be derived using the
identities involving the sine and cosine of a sum or difference of two
numbers.

PRODUCT IDENTITIES

PRODUCT IDENTITIES Examples
Express the following products as a sum.
(a) cos 5x cos 2x
(b) sin 6x sin 2x
(c) sin 5x cos 3x
(d) cos 5x sin 7x

PRODUCT IDENTITIES
The sum identities can be derived using the identities involving
the sine and cosine of a sum or difference of two numbers. Consider
equations (5) to (8).
If we let 𝛼 + 𝛽 = 𝑥 and 𝛼 − 𝛽 = 𝑦, and solve for 𝛼 and 𝛽, we get the
equations
𝛼 = 𝑥+𝑦
and 𝛽 = 𝑥−𝑦
.
2 2

PRODUCT IDENTITIES
Hence, we have the following sum identities:

PRODUCT IDENTITIES Examples
Express the given sum or difference as a product:
(a) cos 4x + cos 2x
(b) cos 5x − cos 3x
(c) sin 6x + sin 2x
(d) sin 3x − sin x

PROVING A GIVEN IDENTITY
An identity is any equation which is true for all replacement
values of the argument of the circular functions. To prove an identity
means to make the left-hand and right-hand members of the equation
identical. This can be done by manipulating one side and transforming
it into an expression found on the other side of the equation. In
instances, when the two sides seem to be equally complicated, both
sides can be manipulated at the same time and reduced to simpler
expressions that are equal.
There is no single method that will help us establish identities.
We have to practice a lot for us to acquire the skill.

PROVING A GIVEN IDENTITY Examples

UNIT 6: Inverse
Trigonometric
Functions

LEARNING OBJECTIVES
I⭲:císc ľíigo⭲omctíic Ïu⭲ctio⭲s
1. Define inverse circular relations;
2. Define inverse circular functions;
3. State the domain and range of inverse circular functions;
4. Evaluate values of inverse circular functions; and
5. Solve related problems involving inverse circular
functions.

INVERSE SINE FUNCTION
In the previous lessons, we were concerned about the function
values of the six trigonometric functions. The trigonometric function
of an angle or a real number, if it exists, is unique. Thus, if x is any
angle in standard position or if x is any real number, then

Now, suppose y is any real number in the closed interval [-1, 1].
We want to find the value of x so that sin x = y. This value of x is
called the inverse sine value of y or arcsin(y). We can raise a similar
problem involving the cosine and the tangent functions.

In this definition, it is important to note that the values of the
arcsine function are restricted to the numbers in the closed interval
1 1
2 2
− 𝜋, 𝜋 or angles in the closed interval [−90°, 90°]. These
intervals define angles whose terminal sides lie in either QI or QIV.

INVERSE SINE FUNCTION Examples
Evaluate the following expressions:
(a) sin−1(1)
1
2
3
(b)sin−1
(c)sin−1 −
(d)sin−1 −
2
2
2

Solution:

I⭲:císc ľíigo⭲omctíic Ïu⭲ctio⭲s Hepolito,Remaík O.,11/24/2021 |Pagc G«
Solution:

INVERSE COSINE FUNCTION
I⭲:císc ľíigo⭲omctíic Ïu⭲ctio⭲s Hepolito,Remaík O.,11/24/2021 |Pagc G4

INVERSE COSINE FUNCTION Examples
(a) cos−1(1)
(b)cos−1
(c)cos−1
(d)cos−1
2
2
− 1
2
3
2

Solution:
I⭲:císc ľíigo⭲omctíic Ïu⭲ctio⭲s Hepolito,Remaík O.,11/24/2021 |Pagc GG

Solution:

INVERSE TANGENT FUNCTION
I⭲:císc ľíigo⭲omctíic Ïu⭲ctio⭲s Hepolito,Remaík O.,11/24/2021 |Pagc G®

INVERSE TANGENT FUNCTION Examples
(a) tan−1(1)
− 1
3
(b)tan−1 3
(c)tan−1
(d)tan−1 −1

Solution:
I⭲:císc ľíigo⭲omctíic Ïu⭲ctio⭲s Hepolito,Remaík O.,11/24/2021 |Pagc 70

Solution:

UNIT 7:
TRIGONOMETRIC
EQUATIONS
11/17/21

LEARNING OBJECTIVES
ľíigo⭲omctíic Equatio⭲s Hepolito,Remaík O.,11/24/2021 |Pagc 7«
1. Solve trigonometric equations involving linear trigonometric
expressions;
2. Solve trigonometric equations involving multiple angles; and
3. Solve trigonometric equations involving quadratic
trigonometric expressions.

CONDITIONAL TRIGONOMETRIC EQUATIONS
We learned about the different identities which are equations
satisfied by all permissible values of real numbers. Examples of these are
the Pythagorean Identities such as sin2 𝑥 + cos2 𝑥 = 1;
1 + tan2 𝑥 = sec2 𝑥; and 1 + 𝑐𝑜𝑡2 𝑥 = csc2 𝑥. In this section, we will
consider trigonometric equations which are satisfied only by some real
numbers. We call these equations conditional trigonometric equations.
Since the sine and cosine functions are periodic, if we can solve
equations involving these functions in the interval [0, 2 𝜋], we can make
use of the period of the function to obtain all solutions. We define the
solutions in the interval [0, 2 𝜋] as fundamental solutions of the
equation.
ľíigo⭲omctíic Equatio⭲s Hepolito,Remaík O.,11/24/2021 |Pagc 74

CONDITIONAL TRIGONOMETRIC EQUATIONS
We now give a series of examples to illustrate how to solve
other conditional trigonometric equations. The methods for solving
conditional trigonometric equations are the same as in ordinary
algebraic equations.
In the process of solving the equation, we use the same
principles of equality to obtain an equivalent equation. These
principles are the addition and multiplication principles of equality
which translate to the usual transportation rules for solving
equations.
Knowledge of the different identities will also be useful in
obtaining equivalent equations.

Example
ľíigo⭲omctíic Equatio⭲s Hepolito,Remaík O.,11/24/2021 |Pagc 7G
Solve the following equationns for 0 ≤ 𝑥 ≤
1
2
𝜋
(a) sin 𝑥 − 1 = 0
(b) sin 𝑥 cos 𝑥 = 0

Solution

Solution
ľíigo⭲omctíic Equatio⭲s Hepolito,Remaík O.,11/24/2021 |Pagc 7®

QUADRATIC TRIGONOMETRIC EQUATIONS
We now consider quadratic trigonometric equations. To solve
such an equation, we can use the square root method (when there is
no linear trigonometric term), or method of factoring or by using the
quadratic formula.
The following examples illustrate these ideas.

Example
Find the solution set of the given equation if t belongs to the half
open interval [0, 2𝜋).
(a) 4sin2 𝑡 − 1 = 0
(b) sec2 𝑡 − 2 tan 𝑡 = 0
(c) 2cos2 𝑡 + 3 cos 𝑡 = 0
ľíigo⭲omctíic Equatio⭲s Hepolito,Remaík O.,11/24/2021 |Pagc ®0

Solution

Solution
ľíigo⭲omctíic Equatio⭲s Hepolito,Remaík O.,11/24/2021 |Pagc ®«

Trigonometric Equations Involving Multiple
Angles Example
Find the fundamental solutions of the following equations.
(a) 𝑠𝑖𝑛 2𝑥 = 0
1
(b) 𝑐𝑜𝑠 𝑥 = −
1
2
(c) tan
2
2 1
2
𝑥 = 3

Solution

Solution
ľíigo⭲omctíic Equatio⭲s Hepolito,Remaík O.,11/24/2021 |Pagc ®G

Solution

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