Trigonometric Identities 2 2 1 sin x cos x sinx tan x cos x cos x cot x sinx 2 2 1 tan x sec x 2 2 1 cot x cosec x
Trigonometric Identities2 2 1sin x cos x
sin xtan x
cos x
cos xcot x
sin x
2 21 tan x sec x
2 21 cot x cosec x
ExamplesProve that (1 – cos A)(1 + sec A) sin A tan A
L.H.S. (1 – cos A)(1 + sec A) = 1 + sec A – cos A –Cos A sec A = 1 + sec A – cos A - 1
= sec A – cos A
1cos A
cos A
21 cos A
cos A
2sin A
cos A
sin Asin A
cos A
= sin A tan A= R.H.S.
ExamplesProve that cot A + tan A sec A cosec A
L.H.S.
2 2cos A sin A
sin Acos A
1
sin Acos A
cos ecAsec A = R.H.S.
cos A sin Acot A tan A
sin A cos A
Examples
R.H.S.
2
1
1 sin
2
1
cos
2sec = R.H.S.
1
1cos ec sin
cos ec sin sinsin
2 cos ecPr ovethat sec
cos ec sin
Solving equationsSolve 2 tan2 x – 7 sec x + 8 = 0 for 0 x 360
2 (sec2x – 1) – 7 sec x + 8 = 0 2 sec2x – 2 – 7 sec x + 8 = 0
2 sec2x – 7 sec x + 6 = 0
(2 sec x – 3)(sec x – 2)= 0
sec x = 3/2 or sec x = 2
cos x = 2/3 or cos x = ½
x = 48.2 or x = 60
or: x = 360 – 48.2 or x = 360 - 60
complete solution: x = 48.2 or 60 or 300 or 311.8
Solving equationsSolve 2 cos x = cot x for 0 x 360
2 cos x = cos x/ sin x
2 cos x sin x = cos x
2 cos x sin x – cos x= 0
cos x(2 sin x – 1)= 0
cos x = 0 or sin x = ½
cos x = 0 x = 90 or 270
sin x = ½ x = 30 or 330
complete solution: x = 30 or 90 or 270 or 30
Solving equationsSolve 3 cot2 x – 10 cot x + 3 = 0 for 0 x 2
(3 cot x - 1)(cot x – 3) = 0
cot x = 1/3 or cot x = 3
tan x = 3 or tan x = 1/3
tan x = 3 x = 1.24c or 4.39c
tan x = 1/3 x = 0.32c or 3.46c
complete solution: x = 0.32c or 1.24c or 3.46c or 4.39c
Solving equationsSolve 5 cot2 x – 2 cosec x + 2 = 0 for 0 x 2
5(cosec2 x – 1) – 2 cosec x + 2 = 0
5cosec2 x – 5 – 2 cosec x + 2 = 0
5cosec2 x – 2 cosec x - 3 = 0
22
5 23 5 2 3 0sin x sin x
sin x sin x
23 2 5 0sin x sin x
3 5 1 0( sin x )(sin x )
sin x = -5/3 not possible or sin x = 1 x = /2
Additional formulaesin (A + B) = sin A cos B + sin B cos Asin (A - B) = sin A cos B - sin B cos Acos (A + B) = cos A cos B - sin A sin Bcos (A - B) = cos A cos B + sin A sin B
1
tan A tan Btan( A B )
tan Atan B
1
tan A tan Btan( A B )
tan Atan B
ExamplesFind the exact value of sin 75
sin (A + B) = sin A cos B + sin B cos Asin (30 + 45) = sin 30 cos 45 + sin 45 cos 30
1 2 2 3
2 2 2 2
2 6 2 6
4 4 4
ExamplesExpress cos (x + /3) in terms of cos x and sin x
1 3
2 2cos x sin x
cos (A + B) = cos A cos B - sin A sin B
cos (x + /3) = cos x cos /3 - sin /3 sin x
Examplessin( A B )
Pr ovethat tan A tan Bcos Acos B
L.H.S.sin A sin B
tan A tan Bcos A cos B
sin Acos B sin B cos A
cos Acos B
sin( A B )
cos Acos B
=
R.H.S.
Double angle formulaesin (A + B) = sin A cos B + sin B cos Asin (A + A) = sin A cos A + sin A cos A sin 2A = 2 sin A cos A
cos (A + B) = cos A cos B - sin A sin Bcos (A + A) = cos A cos A- sin A sin Acos (A + A) = cos2A - sin2A
cos 2A = cos2A - sin2A
cos 2A = 2cos2A - 1
cos 2A = 1 – 2sin2A
Double angle formulae
1
tan A tan Btan( A B )
tan Atan B
1
tan A tan Atan( A A )
tan Atan A
2
22
1
tan Atan A
tan A
2 2Pr ovethat tan A tan A tan Asec A 2 2
2 2 2
2 2 1 1
1 1 1
tan A tan A tan A( tan A ) tan A( tan A )tan A
tan A tan A tan A
2
22
1 2
tan Asec A tan Atan Asec A
tan A cos A
ExamplesGiven that cos A = 2/3, find the exact value of cos 2A.
cos 2A = 2cos2A - 12
22 1
3
8 11
9 9
Given that sin A = ¼ , find the exact value of sin 2A.
sin 2A = 2 sin A cos A
1 15 152
4 4 8 A
4 1
15
Solving equationsSolve cos 2A + 3 + 4 cos A = 0 for 0 x 2
=2 cos2A - 1+ 3 + 4 cos A = 0
=2 cos2A + 4 cos A + 2= 0
= cos2A + 2 cos A + 1 = 0
= cos2A + 2 cos A + 1 = 0
= (cos A + 1)2 = 0
= cos A = - 1
A =
Solving equationsSolve sin 2A = sin A for - x
=2sin A cos A = sin A
=2 sin A cos A – sin A = 0
= sin A(2 cos A – 1) = 0
sin A = 0 or cos A = ½
sin A = 0 A = - or 0 or
cos A = ½ A = - /3 or /3
Complete solution: A = - or - /3 or 0 or /3 or
Solving equationsSolve tan 2A + 5 tan A = 0 for 0 x 2
Complete solution: A= 0.97 , 2.27, 4.01, 5.41c 0, or 2
22
25 2 5 1 0
1
tan Atan A tan A tan A( tan A )
tan A
22 5 1 0tan A[ ( tan A )]
22 5 5 0tan A[ tan A )
27 5 0tan A[ tan A )
tan A = 0 A = 0 or or 2
7 – 5tan2 A = 0 tan A = 7/5 A = 0.97 , 2.27, 4.01 or 5.41c
Harmonic form
If a and b are positive a sin x + b cos x can be written in the form R sin( x + )
a cos x + b sin x can be written in the form R cos( x - )
a sin x - b cos x can be written in the form R sin( x - )
a cos x - b sin x can be written in the form R cos( x + )
2 2R a b
R cos a and R sin b
ExamplesExpress 3 cos x + 4 sin x in the form R cos( x - )
R cos( x - ) = R cos x cos + R sin x sin
3 cos x + 4 sin x = R cos x cos + R sin x sin
R cos = 3 [1] R sin = 4 [2]
[1]2 + [2]2 : R2 sin2 x + R2 cos2 x = 32 + 42
R2(sin2 x + cos2 x ) = 32 + 42
R2= 32 + 42 = 25 R = 5
[2] [1]: tan = 4/3 = 53.1
3 cos x + 4 sin x = 5 cos( x + 53.1 )
ExamplesExpress 12 cos x + 5 sin x in the form R sin( x + )
R sin( x + ) = R sin x cos + R cos x sin
12 cos x + 5 sin x = R sin x cos + R cos x sin
R cos = 12 [1] R sin = 5 [2]
[1]2 + [2]2 : R2 cos2 x + R2 sin2 x = 122 + 52
R2(cos2 x + sin2 x ) = 122 + 52
R2= 122 + 52 = 169 R = 13
[2] [1]: tan = 5/12 = 22.6
12 cos x + 5 sin x = 13 sin( x + 22.6 )
ExamplesExpress cos x - 3 sin x in the form R cos( x + )
R cos( x + ) = R cos x cos - R sin x sin
cos x - 3 sin x = R cos x cos - R sin x sin
R cos = 1 [1] R sin = 3 [2]
[1]2 + [2]2 : R2 cos2 x + R2 sin2 x = 12 + (3 ) 2
R2(cos2 x + sin2 x ) = 12 + 3
R2= 1 + 3 = 4 R = 2
[2] [1]: tan = 3 = 60
cos x + 3 sin x = 2 cos( x + 60 )
Solving equationsSolve 7 sin x + 3 cos x = 6 for 0 x 2
R sin( x + ) = R sin x cos + R cos x sin
7 sin x + 3 cos x = R sin x cos + R cos x sin
R cos = 7 [1] R sin = 3 [2]
R2 = 72 + 32 R = 7.62
[2] [1]: tan = 3/7 = 0.405c (Radians)
7 sin x + 3 cos x = 7.62 sin( x + 0.405)
7.62 sin( x + 0.405 ) = 6 x + 0.405 = sin-1(6/7.62)
x + 0.405 = 0.907 or 2.235
x = 0.502c or 1.830c