Double angle identities 1.sin2θ =2sinθcosθ 2... Useful suggestions for proving trigonometric identities 1.. Where possible, express different functions in terms of the same function..
Trang 1Trigonometric Identities and Equations
I Fundamental Trigonometric Identities
A Reciprocal identities
1
θ
θ
cos
1
θ
θ sin
1
θ
θ tan
1
B.Quotient identities
1
θ
θ θ
cos
sin
θ
θ θ
sin
cos cot =
C Pythagorean identities
1 sin2θ +cos2θ =1 2 2θ 2θ 3
sec 1
csc cot
D Sum and difference identities
1 sin(θ ±φ)=sinθcosφ±cosθsinφ
2 cos(θ ±φ)=cosθcosφmsinθsinφ
3
φ θ
φ θ
φ
θ
tan tan 1
tan tan
) tan(
m
±
=
±
E Double angle identities
1.sin2θ =2sinθcosθ
2 cos2θ =cos2θ −sin2θ =1−2sin2θ =2cos2θ −1
3
θ
θ θ
2 tan 1
tan 2 2
tan
−
=
F Half angle identities
1
2
cos 1 2 sin2 θ = − θ
⎟
⎠
⎞
⎜
⎝
⎛
2
2
cos 1 2 cos2 θ = + θ
⎟
⎠
⎞
⎜
⎝
⎛
3
θ
θ θ
cos 1
cos 1 2
tan2
+
−
=
⎟
⎠
⎞
⎜
⎝
⎛
Trang 2G Miscellaneous identities
1 sin(−θ)=−sinθ
2 cos(−θ)=cosθ
3 tan(−θ)=−tanθ
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛ ±
=
±
2
cos 2
sin 2 sin
⎠
⎞
⎜
⎝
⎛ −
⎟
⎠
⎞
⎜
⎝
⎛ +
= +
2
cos 2 cos 2 cos
⎠
⎞
⎜
⎝
⎛ −
⎟
⎠
⎞
⎜
⎝
⎛ +
−
=
−
2
sin 2 sin 2 cos
2
1 ) sin(
2
1 cos
sinθ ϕ = θ +ϕ + θ −ϕ
2
1 ) sin(
2
1 sin
cosθ ϕ = θ +ϕ − θ −ϕ
2
1 ) cos(
2
1 cos
cosθ ϕ = θ +ϕ + θ −ϕ
2
1 ) cos(
2
1 sin
sinθ ϕ = θ −ϕ − θ +ϕ
11
θ
θ θ
θ θ
sin
cos 1 cos 1
sin 2
= +
=
⎟
⎠
⎞
⎜
⎝
⎛
H Useful suggestions for proving trigonometric identities
1 Avoid aimless transformations Any transformation that is made in one of the members should lead in some way to the form of the other
2 Start with the more complicated member of the identity and transform it into the form of the simpler member
3 Where possible, express different functions in terms of the same function
4 It is often useful to express all functions in terms of sines and cosines, or in terms of tangents and secants
Trang 35 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
1 Using trigonometric identities and fundamental trigonometric function values, find each of the following:
(a)
2
3 2 4
3 2 2
2
3 1 2
60 cos 1 2
30 sin 15
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
o
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
−
= +
=
2
3 2
2 30
sin 45 sin 30 cos 45 cos ) 30 45 cos(
75
4
2 6 2
1 2
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
(c)
2
1 30 sin )]
15 ( 2 sin[
15 cos 15 sin
3 2 1 2
3 1 2 1 30
cos 1
30 sin 2
30 tan 15
+
= +
= +
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
o
o o
o
2
1 ) 5 7 5 37 cos(
2
1 5 7 cos 5 37
4
3 2 2
3 2
1 2
2 2
1 30 cos 2
1 45 cos 2
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛ +
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=
o
Trang 4
2 Prove: x x
x
x
tan sin
csc
sec
x x
x
x x
x
x x
x
x
sin 1
sin cos 1 sin sin
1 cos
1 sin csc
sec csc
1 csc
sec 1
x
x
tan sin
cos
sin
+
=
3 Prove:
x x
x x
x
2 sin
2 cos
sin sin
cos
= +
x x x
x
x x
x x
x x x
x
x x x
x x
x
cos sin
1 cos
sin
sin cos
cos sin
) (sin sin cos
sin
) (cos cos cos
sin sin
x x
x sin2
2 cos
sin 2
2
=
4 Prove:
y x
y x
y x
y x
tan tan
tan tan
) sin(
) sin(
−
+
=
− +
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
=
−
+
=
−
+
) )(cos (cos cos
sin cos
sin
) )(cos (cos cos
sin cos
sin
cos
sin cos
sin
cos
sin cos
sin
tan tan
tan tan
y x y
y x
x
y x y
y x
x
y
y x
x
y
y x
x y
x
y x
) sin(
) sin(
cos sin cos sin
cos sin cos
sin
y x
y x x
y y
x
x y y
x
−
+
=
− +
II Solution of Trigonometric Equations
A Useful suggestions for solving trigonometric equations
1 Simplify the equation by clearing fractions, removing parentheses, combining like terms, and removing radicals
2 Express functions of a double angle, a half angle, or the sums and differences
of angles in terms of functions of the single angle; then express the different functions of the single angle in terms of a single function of that angle
Trang 53 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
1 Solve for x in the interval [0,2π ): 2sinx+ 3=0
2
3 sin
3 sin
2 0 3 sin
x x
3 2
3
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
Since sinx is negative, x lies in the 3rd and 4th
quadrants Thus, (240 )
3
4 ) 60 ( 3 ) 180
=
(300 )
3
5 ) 60
(
3
o
Both of these values do check
2 Solve for x in the interval [0,2π ): cosx=cosxtanx
cosx=cosxtanx⇒ cosx−cosxtanx=0⇒cosx(1−tanx)=0⇒
cosx=0 or 1−tanx=0⇒cosx=0 or tanx=1
2 0
=
⇒
x (90o)
or (270 )
2
=
x tan x= 1⇒ reference angle = (45 )
4 ) 1 (
=
tanx is positive, x lies in the 1st and 3rd quadrants Thus, (45 )
4 o π
=
4
5 ) 45 ( 4 ) 180
2
3 , 4
5 , 2
, 4
π π π π
or
x= and they all check
3 Solve for x in the interval [0,2π ): 2cos3x=1
2
1 3 cos 1 3 cos
2 x= ⇒ x= Since 0≤ x<2π , 0≤ x<6π The reference angle
for 3x is (60 )
3 2
1
⎟
⎠
⎞
⎜
⎝
⎛
− and cosx is positive in the 1st
and 4th quadrants
Thus, (60 )
3
3 =π o
3
5 ) 60 ( 3 ) 360 ( 2
x , 3x=2π(360o)+
Trang 6(420 )
3
7 ) 60
(
3
o
3
11 ) 60 ( 3 ) 720 ( 4
3
13 ) 60
(
3
o
3
17 ) 60 ( 3 ) 1080 ( 6
x
9
17 , 9
13 , 9
11 , 9
7 , 9
5 , 9
π π
π π π π
or
4 Solve for x in the interval [0,2π ): cos2x=cosx
cos2x=cosx⇒cos2x−cosx=0⇒(2cos2 x−1)−cosx=0⇒2cos2 x− cosx−1=0⇒(2cosx+1)(cosx−1)=0⇒ 2cosx+1=0or cos x−1=0⇒
2
1 cosx=− or cosx=1 =− ⇒
2
1
cos x reference angle is
3 2
1 cos 1 =π
⎟
⎠
⎞
⎜
⎝
⎛
−
and
x lies in the 2nd or 3rd quadrants since cosx is negative
3
2 3
π π
π − =
=
3
4 3
π π
π + =
=
x cosx=1⇒ x=0 Thus,
3
4 , 3
2 ,
or
x= and they all check
5 Solve for x in the interval [0,2π ): sinx=cosx
= ⇒ =1⇒tan =1⇒
cos
sin cos
x
x x
4 ) 1 ( tan−1 =π
and
x lies in the 1st or 3rd quadrants since tanxis positive
4
π
=
⇒ x or x=π +
4
5 4
π
π =
and they both check
Trang 7Practice Sheet – Trigonometric Identities and Equations
I Verify the following identities:
x
x
tan sin
csc
sec
x
x x
2 sin 1
2 cos 4
tan
−
=
⎟
⎠
⎞
⎜
⎝
⎛ +π
(3)
x
x x
sin 2
sec 2
x
x x
sec 2
1 sec 2
⎟
⎠
⎞
⎜
⎝
⎛
x x
x x
3 tan 4
sin 2
sin
2 cos 4
−
−
II Solve the following equations for all values of x in the interval [0,2π ):
(1) 3sinx−4=5sinx−3 (2) 2sinxcosx= 3cosx
(3) 4cos2 x−1=0 (4) 2cos2 x+3sinx=3
(5) sinx−cosx=1
Solution Key for Trigonometric Identities and Equations
x x
x
x x
x
x x
x
x
tan sin
1
sin cos
1 sin
sin 1 cos
1 sin csc
sec csc
1 csc
sec
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛ +
= +
= +
= +
−
+
=
−
+
=
−
+
=
−
+
=
⎟
⎠
⎞
⎜
⎝
⎛ +
x x
x x x
x x x x
x x
x x
sin cos
sin cos cos
sin 1 cos
sin 1 tan 1
tan 1 tan 4 tan 1
tan 4 tan 4
π π
x
x x
x x x
x x
x x x x
x x x x
2 sin 1
2 cos sin
sin cos 2 cos
sin cos
) sin )(cos sin (cos
) sin )(cos sin (cos
2 2
2 2
−
= +
−
−
=
−
−
− +
Trang 8(3)
x
x x
x x
x x
x x
x
sin 2
sec sec
sin 2
1 cos
1 sin 2
1 cos
sin 2
1 2
sin
1 2
(4)
x
x x
x
x x
x x
x x
x
sec 2
1 sec sec
2 cos
cos cos
1
cos 2 cos
cos 1 2
cos 1 2
⎟
⎠
⎞
⎜
⎝
⎛
−
−
=
⎟
⎠
⎞
⎜
⎝
⎟
⎠
⎞
⎜
⎝
⎟
⎠
⎞
⎜
⎝
⎟
⎠
⎞
⎜
⎝
−
=
−
−
x x
x x x
x x
x
x x x
x x
x
x x
3 cos ) sin(
2
sin 3 sin 2 2
4 2 cos 2
4 2 sin 2
2
2 4 sin 2
2 4 sin 2 4
sin 2
sin
2 cos 4
cos
x
x x
x
x x
3 tan 3
cos
3 sin 3
cos sin
2
sin 3 sin
−
−
II (1)
6
11 , 6
7 2
1 sin
3 sin 5 4 sin
x x
x
(2)
2
3 sin
cos 3 cos
sin
3
2 , 3 0
cos = ⇒ =π π
x
2
3 , 2
π π
=
x
and they all check
(3)
3
5 , 3
4 , 3
2 , 3 2
1 cos
0 1 cos
x x
(4) 2cos2 x+3sinx=3⇒2(1−sin2 x)+3sinx=3⇒2sin2 x−3sinx+1=0⇒
2
1 sin 0 ) 1 )(sin 1 sin
2
6
5 , 6 1
sin = ⇒ =π π
x
2
π
=
x and they all check
(5) sinx−cosx=1⇒sinx=1+cosx⇒(sinx) (2 = 1+cosx)2 ⇒sin2 x=1+
2cosx+cos2 x⇒1−cos2 x=1+2cosx+cos2 x⇒ 2cos2 x+2cosx=0⇒
2cosx(cosx+1)=0⇒cosx=0 or
2
3 , 2 1
cos =− ⇒ =π π
x
x or x=π
2
π
=
x or x=π are the solutions because they both check However,
2
3π
=
x
does not check in the original equation and thus is not a solution [Note:
2
3π
=
x is an extraneous root created by squaring both sides of the original
Trang 98