Trigonometric Identities and Equations
Trigonometric Identities and Equations
Trigonometric Identities and Equations
A. Reciprocal identities
1 1 1
1. secθ = 2. cscθ = 3. cot θ =
cos θ sin θ tan θ
B.Quotient identities
sin θ cos θ
1. tan θ = 2. . cot θ =
cos θ sin θ
C. Pythagorean identities
tan θ ± tan φ
3. tan(θ ± φ ) =
1 m tan θ tan φ
2 tan θ
3. tan 2θ =
1 − tan 2 θ
1
G. Miscellaneous identities
1. sin( −θ ) = − sin θ
2. cos( −θ ) = cos θ
3. tan( −θ ) = − tan θ
⎛θ ± ϕ ⎞ ⎛θ m ϕ ⎞
4. sin θ ± sin ϕ = 2 sin ⎜ ⎟ cos⎜ ⎟
⎝ 2 ⎠ ⎝ 2 ⎠
⎛θ + ϕ ⎞ ⎛θ − ϕ ⎞
5. cos θ + cos ϕ = 2 cos⎜ ⎟ cos⎜ ⎟
⎝ 2 ⎠ ⎝ 2 ⎠
⎛θ + ϕ ⎞ ⎛θ − ϕ ⎞
6. cos θ − cos ϕ = −2 sin ⎜ ⎟ sin ⎜ ⎟
⎝ 2 ⎠ ⎝ 2 ⎠
1 1
7. sin θ cos ϕ = sin(θ + ϕ ) + sin(θ − ϕ )
2 2
1 1
8. cos θ sin ϕ = sin(θ + ϕ ) − sin(θ − ϕ )
2 2
1 1
9. cos θ cos ϕ = cos(θ + ϕ ) + cos(θ − ϕ )
2 2
1 1
10. sin θ sin ϕ = cos(θ − ϕ ) − cos(θ + ϕ )
2 2
⎛θ ⎞ sin θ 1 − cos θ
11. tan ⎜ ⎟ = =
⎝ 2 ⎠ 1 + cos θ sin θ
2
5. As a rule, trigonometric functions of a double angle, a half angle, or the sums
and differences of angles should be expressed in terms of functions of the
single angle.
6. Simplify expressions by utilizing basic identities and combining like terms.
7. Simplify fractions. For example, transform complex fractions into simple
fractions or divide the terms of a fraction by the common factors.
I. Examples
⎛ 30 o ⎞ 1 − cos 60 o 1− 3 2− 3 2− 3
o
(a) sin 15 = sin ⎜ ⎟ = = 2 = =
⎜ 2 ⎟ 2 2 4 2
⎝ ⎠
⎛ 2 ⎞⎛ 3 ⎞
(b) cos 75o = cos( 45o + 30 o ) = cos 45 o cos 30 o − sin 45 o sin 30 o = ⎜⎜ ⎟⎜ ⎟
⎟⎜ 2 ⎟ −
⎝ 2 ⎠⎝ ⎠
⎛ 2 ⎞⎛ 1 ⎞ 6− 2
⎜ ⎟
⎜ 2 ⎟⎜⎝ 2 ⎟⎠ = 4
⎝ ⎠
1
(c) 2 sin 15 o cos 15o = sin[ 2(15o )] = sin 30 o =
2
⎛ 30 o ⎞ o 1
(d) tan 15 = tan ⎜
o ⎟ = sin 30 = 2 =
1
= 2− 3
⎜ 2 ⎟ 1 + cos 30 o 2+ 3
⎝ ⎠ 1+ 3
2
1 1
(e) cos 37.5o cos 7.5o = cos(37.5o + 7.5o ) + cos(37.5o − 7.5o ) =
2 2
1 1 ⎛ 1 ⎞⎛ 2 ⎞⎟ ⎛ 1 ⎞⎛⎜ 3 ⎞⎟ 2+ 3
cos 45 o + cos 30 o = ⎜ ⎟⎜⎜ +⎜ ⎟ =
2 2 ⎝ 2 ⎠⎝ 2 ⎟⎠ ⎝ 2 ⎠⎜⎝ 2 ⎟⎠ 4
3
1 + sec x
2. Prove: = sin x + tan x
csc x
( )( )
1
1 + sec x 1 sec x
= + = sin x + cos x = sin x + 1 sin x = sin x +
csc x csc x csc x 1 cos x 1
sin x
sin x
= sin x + tan x
cos x
cos x sin x 2
3. Prove: + =
sin x cos x sin 2 x
2 2
=
2 sin x cos x sin 2 x
4
3. Solve the resulting equation, whether it be linear or quadratic in nature, for all
the values of the angle in the given domain.
4. Checks the results by substituting into the original equation.
B. Examples
− 3
2 sin x + 3 = 0 ⇒ 2 sin x = − 3 ⇒ sin x = . Get the reference angle
2
⎛ 3⎞ π
sin −1 ⎜⎜ ⎟ = (60 o ). Since sin x is negative, x lies in the 3rd and 4th
⎟
⎝ 2 ⎠ 3
π 4π
quadrants. Thus, x = π (180 o ) + (60 o ) = (240 o ) or x = 2π (360o ) −
3 3
π 5 π
(60 o ) = (300 o ) . Both of these values do check.
3 3
1
2 cos 3 x = 1 ⇒ cos 3 x = . Since 0 ≤ x < 2π , 0 ≤ x < 6π . The reference angle
2
⎛1⎞ π
for 3x is cos −1 ⎜ ⎟ = (60 o ) and cos x is positive in the 1st and 4th quadrants.
⎝2⎠ 3
π π 5π
Thus, 3 x = (60 o ) , 3 x = 2π (360 o ) − (60 o ) = (300 o ) , 3 x = 2π (360 o ) +
3 3 3
5
π 7π π 11π
(60 o ) = (420 o ) , 3 x = 4π (720 o ) − (60 o ) = (660 o ) , 3 x = 4π (720 o ) +
3 3 3 3
π 13π π 17π
(60 o ) = (780 o ) , and 3 x = 6π (1080 o ) − (60 o ) = (1020 o ) ⇒
3 3 3 3
1 1 ⎛1⎞ π
cos x = − or cos x = 1 . cos x = − ⇒ reference angle is cos −1 ⎜ ⎟ = and
2 2 ⎝2⎠ 3
π 2π
x lies in the 2nd or 3rd quadrants since cos x is negative ⇒ x = π − = or
3 3
π 4π 2π 4π
x =π + = . cos x = 1 ⇒ x = 0 . Thus, x = 0, , or and they all
3 3 3 3
check.
sin x π
sin x = cos x ⇒ = 1 ⇒ tan x = 1 ⇒ reference angle is tan −1 (1) = and
cos x 4
π
x lies in the 1st or 3rd quadrants since tan x is positive ⇒ x = or x = π +
4
π 5π
= and they both check.
4 4
6
Practice Sheet – Trigonometric Identities and Equations
I. Verify the following identities:
1 + sec x ⎛π ⎞ cos 2 x
(1) = sin x + tan x (2) tan ⎜ + x ⎟ =
csc x ⎝4 ⎠ 1 − sin 2 x
sec x ⎛ x ⎞ sec x − 1
(3) csc 2 x = (4) sin 2 ⎜ ⎟ =
2 sin x ⎝ 2 ⎠ 2 sec x
cos 4 x − cos 2 x
(5) = tan 3 x
sin 2 x − sin 4 x
II. Solve the following equations for all values of x in the interval [0, 2π ) :
1
1 + sec x 1 sec x ⎛ 1 ⎞⎛ sin x ⎞
I. (1) = + = sin x + cos x = sin x + ⎜ ⎟⎜ ⎟ = sin x + tan x
csc x csc x csc x 1 ⎝ cos x ⎠⎝ 1 ⎠
sin x
⎛π ⎞
tan⎜ + x ⎟ =
(
tan π
4
) + tan x
=
1 + tan x
=
1+
sin x
cos x = cos x + sin x =
⎠ 1 − tan (π )tan x 1 − tan x 1 − sin x cos x − sin x
(2)
⎝4
4
cos x
7
1 1 1 1 1 sec x
(3) csc 2 x = = = ⋅ = ⋅ sec x =
sin 2 x 2 sin x cos x 2 sin x cos x 2 sin x 2 sin x
1 − cos x 1 cos x
−
⎛ x ⎞ 1 − cos x sec x − 1
(4) sin 2 ⎜ ⎟ = = cos x = cos x cos x =
⎝2⎠ 2 2 2 sec x 2 sec x
cos x
⎛ 4x + 2x ⎞ ⎛ 4x − 2x ⎞
− 2 sin ⎜ ⎟ sin ⎜ ⎟
cos 4 x − cos 2 x ⎝ 2 ⎠ ⎝ 2 ⎠ − 2 sin 3x sin x
(5) = = =
sin 2 x − sin 4 x ⎛ 2x − 4x ⎞ ⎛ 2x + 4x ⎞ 2 sin( − x) cos 3x
2 sin ⎜ ⎟ cos⎜ ⎟
⎝ 2 ⎠ ⎝ 2 ⎠
− 2 sin 3 x sin x sin 3 x
= = tan 3 x
− 2 sin x cos 3 x cos 3 x
1 7π 11π
II. (1) 3 sin x − 4 = 5 sin x − 3 ⇒ sin x = − ⇒x= , and they both check.
2 6 6
3 π 2π π 3π
(2) 2 sin x cos x = 3 cos x ⇒ sin x = or cos x = 0 ⇒ x = , or x = ,
2 3 3 2 2
and they all check.
1 π 2π 4π 5π
(3) 4 cos 2 x − 1 = 0 ⇒ cos x = ± ⇒x= , , , and they all check.
2 3 3 3 3