Download Simple Trigonometric Identities and Equations

Further Trigonometric Identities
Exercise 13.1
5.
(f)
tan 5x = tan 2x,
0 < x < 360
sin 5 x
sin 2 x
=
cos 5 x
cos 2 x
sin 5x cos 2x = cos 5x sin 2x
sin 5x cos 2x  cos 5x sin 2x = 0
sin (5x  2x) = 0

sin 3x = 0
0 < 3x < 1080
Basic angle,  = 0
3x = 0, 180, 360, 540, 720, 900, 1080
(reject 0 & 1080)
x = 60, 120, 180, 240, 300
6.
(b)
7.
(e)
Prove:
(g)
Prove: tan A + tan B =
(h)
Prove:
tan 2 x  tan 1
1
= ,
0 < x < 2
1  tan 2 x tan 1
2
1
tan (2x  1) =
1 < 2x  1 < 4  1
2
Basic angle,  = 0.4636
2x  1 = 0.4636,  + 0.4636, 2 + 0.4636, 3 + 0.4636
2x
= 1.4636, 4.6052, 7.7468, 10.888
x = 0.732, 2.30, 3.87, 5.44
sin( A  B)  sin( A  B)
= cot A
cos( A  B)  cos( A  B)
[sin A cos B  cos A sin B]  [sin A cos B  cos A sin B]
L.H.S. =
[cos A cos B  sin A sin B]  [cos A cos B  sin A sin B]
2 cos A sin B
=
 2 sin A sin B
=  cot A
= R.H.S.
sin( A  B )
cos A cos B
sin A cos B  cos A sin B
R.H.S. =
cos A cos B
sin A cos B
cos A sin B
=
+
cos A cos B
cos A cos B
= tan A + tan B
= L.H.S.
sin( A  B)
tan A  tan B
=
cos( A  B) 1  tan A tan B
L.H.S. = [sin A cos B + cos A sin B]  [cos A cos B + sin A sin B]
sin A cos B  cos A sin B
cos A cos B  sin A sin B
=

cos A cos B
cos A cos B
= [tan A + tan B]  [1 + tan A tan B]
tan A  tan B
=
= R.H.S.
1  tan A tan B
Ex 13: Further Trigonometric Identities
Prepared by Mr Ang KS
1
9.
(a)
(b)
(c)
12.
4
3
, sin B =  , and A & B in the same quadrant (3rd quadrant)
y
3
5
tan A  tan 45
tan (A + 45) =
3 A
1  tan A tan 45

4
4
= [ + 1]  [1  ]
4
5
3
3
= 7
cos (A  B) = cos A cos B + sin A sin B
y
3
4
4
3
= ( )( ) + ( )( )
5
5
5
5
4 B
24

=
3
25
5
sin (B  A)
= sin B cos A  cos B sin A
3
3
4
4
= ( )( )  ( )( )
5
5
5
5
7
=
25
Given: tan A =
x
x
Show: sin (A + B) sin (A  B) = sin2 A  sin2 B
L.H.S. = [sin A cos B + cos A sin B]  [sin A cos B  cos A sin B]
= (sin A cos B)2  (cos A sin B)2
= sin2 A (1  sin2 B)  (1  sin2 A)sin2 B
= sin2 A  sin2 A sin2 B  sin2 B + sin2 Asin2 B
= sin2 A  sin2 B
= R.H.S.
(a)
sin2 3x = sin2 x,
0  x  180
sin2 3x  sin2 x = 0
sin (3x + x) sin (3x  x) = 0
sin 4x sin 2x = 0
0  2x  360 or 0  4x  720
sin 4x = 0
or
sin 2x = 0
Basic angle,  = 0
Basic angle,  = 0
4x = 0, 180, 360, 540, 720
2x = 0, 180, 360
x = 0, 45, 90, 135, 180
x = 0, 90, 180
x = 0, 45, 90, 135, 180
(b)
Show: sin 15 sin 75 =
1
4
L.H.S. = sin 75 sin 15
= sin (45 + 30) sin (45  30)
= sin2 45  sin2 30
1
1 2
=(
)  ( )2
2
2
1
1
1
= 
=
= R.H.S.
2
4
4
 End of Ex 13.1 
Ex 13: Further Trigonometric Identities
Prepared by Mr Ang KS
2
Exercise 13.2
2.
(e)
cos 2y  1 = 3sin y,
(1  2sin2 y)  1 = 3sin y
2sin2 y + 3sin y = 0
sin y(2sin y + 3) = 0
0 < y < 360
sin y = 0
sin y = 
Basic angle,  = 0
y = 0, 180, 360
y = 180
3.
or
3
(No solution as 1  sin y  1)
2
(reject 0 & 360, since 0 < y < 360)
(f)
(sin x  cos x)2 = 2,
0 < x < 360
sin2 x + cos2 x  2sin x cos x = 2
1  sin 2x = 2
sin 2x = 1
0 < 2x < 720
Basic angle,  = 90
2x = 180 + 90, 360  90, 540 + 90, 720  90
x = 135, 315
(b)
5sin x cos2 x + 2cos x = 0,
0  x  2, or 0  2x  4
cos x(5sin x cos x + 2) = 0
1
4
cos x = 0
or
5( sin 2x) + 2 = 0, i.e. sin 2x = 
2
5

Basic angle,  =
Basic angle,  = 0.9273
2


x = , 2 
2x =  + 0.9273, 2  0.9273, 3 + 0.9273, 4  0.9273
2
2
 3
x = ,
, 2.03, 2.68, 5.18, 5.82
2 2
(e)
2 tan 2 tan x = 1  tan2 x,
0  x  2, so 0  2x  4
2 tan x
tan 2(
)=1
1  tan 2 x
1
tan 2x =
= 0.4577
tan 2
Basic angle,  = 0.4292
2x =   0.4292, 2  0.4292, 3  0.4292, 4  0.4292
x = 1.36, 2.93, 4.50, 6.07
Ex 13: Further Trigonometric Identities
Prepared by Mr Ang KS
3
(f)
tan 2x = 3tan x,
2 tan x
= 3tan x
1  tan 2 x
0  x  2

2tan x = 3tan x(1  tan2 x)
2tan x  3tan x(1  tan2 x) = 0
tan x [2  3(1  tan2 x)] = 0
tan x(3tan2 x  1) = 0
tan x = 0
tan x = 
or
1
3
Basic angle,  = 0
Basic angle,  =

6




,   ,  + , 2 
6
6
6
6
 5 7 11
x = 0, , 2, ,
,
,
6
6
6 6
x = 0, , 2
6.
x=
Given: cot  = k and  is acute.
1
cot  = k

tan  =
k
(a)
(b)
(d)
sin 2 = 2sin  cos  = 2(
k 2 1
1
1

2k
= 2
k 1
sec 2 = 1  cos 2
k
)(
)
k
k2 1
k2 1
cot ( + 45) = 1  tan ( + 45)
(c)
tan   tan 45
=1
= 1  (1  2sin2 )
1  tan  tan 45
1
1
1
=1 k
= 1  [1  2( 2
)]
1
k

1
1
k
k2 1 2
1 k
k 1
=1[

]
=1
k
k
k2 1
k2 1
1 k
k
=1[

]
= 2
k
k 1
k 1
k 1
k 1
=1
=
k 1
k 1
tan   tan 2
2 tan 
tan 2 =
then, tan 3 =
2
1  tan  tan 2
1  tan 
1
1
1
1
2k
2k
= 2( )  [1  2 ]
=( + 2
)  (1 
 2
)
k
k
k
k
k 1
k 1
k (k 2  1)  2k
k 2 1
2
k 2  1  2k 2
= 
=

k
k2
k (k 2  1)
k (k 2  1)
=
k2
2
 2
k
k 1
=
k (k 2  1)
3k 3  1

k 3  3k
k (k 2  1)
=
2k
2
k 1
=
3k 3  1
k (k 2  3)
Ex 13: Further Trigonometric Identities
Prepared by Mr Ang KS
4
7.
(e)
(f)
9.
sin A
1
= tan A
1  cos A
2
A
A
A
A
2 sin cos
2 sin cos
2
2
2
2
L.H.S. =
=
A
A
1  [2 cos 2  1]
2 cos 2
2
2
A
sin
1
2
=
= tan A
A
2
cos
2
Prove:
= R.H.S.
1  tan 2 A
= sec 2A
1  tan 2 A
sin 2 A
sin 2 A
L.H.S. = [1 +
]

[1

]
cos 2 A
cos 2 A
cos 2 A  sin 2 A
cos 2 A  sin 2 A
=

cos 2 A
cos 2 A
cos 2 A
1
=

cos 2 A cos 2 A
1
=
= sec 2A
= R.H.S.
cos 2 A
Prove:
Given: a = cos x + sin x and b = cos x  sin x
(a)
Prove: cos 2x = ab
L.H.S. = cos2 x  sin2 x
= (cos x + sin x)(cos x  sin x)
= ab
= R.H.S.
(b)
Prove: sin 2x = a2  1
R.H.S. = (cos x + sin x)2  1
= cos2 x + sin2 x + 2sin x cos x  1
= 1 + sin 2x  1
= L.H.S.
(c)
Prove: a2 + b2 = 2
R.H.S. = (cos x + sin x)2 + (cos x  sin x)2
= (cos2 x + sin2 x + 2sin x cos x) + (cos2 x + sin2 x  2sin x cos x)
=1+1
=2
= L.H.S.
 End of Ex 13.2 
Ex 13: Further Trigonometric Identities
Prepared by Mr Ang KS
5
Exercise 13.3
3.
(d)
3  7cos  + 24sin ,
0 <  < 360
Let 24sin   7cos  = Rsin (  )
Then R =
24 2  (7) 2 = 25
and
7
24
so  = 16.26
tan  =
Since 25  25sin (  16.26)  25

3  25  3  7cos  + 24sin   25 + 3
Max. value = 28,
when sin (  16.26) = 1

 = 90 + 16.26 = 106.3
Min. value = 22,
when sin (  16.26) = 1

 = 270 + 16.26 = 286.3
4.
(d)
sin x + 2cos x = 2 ,
Let sin x + 2cos x = Rsin (x + )
0 < x < 360
Then R = 12  2 2 =
5
and
sin x + 2cos x = 2
5 sin (x + 63.43) =
2
2
=2
1
so  = 63.43
tan  =
2
63.43 < x + 63.43 < 423.23
5
Basic angle,  = 39.23
x + 63.43 = 180  39.23, 360 + 39.23
x = 77.3, 335.8
sin (x + 63.43) =
(f)
 cos x + e sin x = 2,
0 < x < 360
Let  cos x + e sin x = Rcos (x  )
Then R =
 2  e 2 = 4.154
 cos x + e sin x = 2
4.154cos (x  40.87) = 2
2
cos (x  40.87) =
4.154
Basic angle,  = 61.22
x  40.87 = 61.22, 360  61.22
x = 102.1, 339.7
Ex 13: Further Trigonometric Identities
Prepared by Mr Ang KS
and
tan  =
e

so  = 40.87
40.87 < x  40.87 < 400.87
6
5.
(b)
2cot x = 3 + 2cosec x,
2 cos x
2
=3+
sin x
sin x
3 sin x  2
2 cos x
=
sin x
sin x
2cos x = 3sin x + 2
2cos x  3sin x = 2
Let 2cos x  3sin x = Rcos (x + )
Then R =
(sin x  0)
2 2  (3) 2 = 13
2cos x  3sin x = 2
13 cos (x + 56.31) = 2
2
cos (x + 56.31) =
13
Basic angle,  = 56.31
x + 56.31 = 56.31, 360  56.31
x = 247.4
8.
0 < x < 360
and
tan  =
so  = 56.31
56.31 < x + 56.31 < 416.31
Given: ORQ = , so
RQX = 90   and PQX = 
PX
In PQX, sin  =
, so PX = 2sin 
2
YR
In QRY, cos  =
, so YR = 8cos 
8
So OR = OY + YR
= PX + YR
= (2sin  + 8cos ) cm
(a)
Let 2sin  + 8cos  = Rsin ( + )
Then R =
(b)
2 2  82 =
68
3
2
and
Q
2 cm
P

X
90  
8 cm
O
Y

R
8
=4
2
so  = 75.96
tan  =
Since  68  68 sin ( + 75.96)  68

 68  OR  68
For max. OR, sin ( + 75.96) = 1

 = 90  75.96 = 14.0
68 sin ( + 75.96) = 6
OR = 6

6
sin ( + 75.96) =
Note that:
68
Basic angle,  = 46.69
so
 + 75.96 = 180  46.69
 = 57.4
x > 0
x + 75.96 > 75.96
 End of Ex 13.3 
Ex 13: Further Trigonometric Identities
Prepared by Mr Ang KS
7
Exercise 13.4
2.
(f)
sin
11
5
 sin
12
12
= 2cos [(
11
5
11
5
+
)  2] sin [(

)  2]
12
12
12
12
2

sin
3
4
1
1
= 2( )(
)
2
2
= 2cos
3.
4.
=
2
2
(e)
cos 3x  2sin 2x = cos x,
0  x  180
cos 3x  cos x  2sin 2x = 0
3x  x
3x  x
2sin
sin
 2sin 2x = 0
2
2
2sin 2x sin x  2sin 2x = 0
2sin 2x (sin x + 1) = 0
sin 2x = 0
or
sin x = 1
Basic angle,  = 0
Basic angle,  = 90
2x = 0, 180, 360
x = 180 + 90, 360  90
x = 0, 90, 180
= 270 (rejected since 0  x  180)
x = 0, 90, 180
(f)
cos 3x  cos x = sin 3x  sin x,
0  x  180
3x  x
3x  x
3x  x
3x  x
2sin
sin
= 2cos
sin
2
2
2
2
2sin 2x sin x = 2cos 2x sin x
2cos 2x sin x + 2sin 2x sinx = 0
2sin x(cos 2x + sin 2x) = 0
sin x = 0
or
sin 2x = cos 2x

tan 2x = 1
Basic angle,  = 0
Basic angle,  = 45
x = 0, 180
2x = 180  45, 360  45
x = 0, 67.5, 157.5, 180
(d)
sin 4x + sin 2x + cos 4x = cos 2x,
0x
(sin 4x + sin 2x) + (cos 4x + cos 2x) = 0
4x  2x
4x  2x
4x  2x
4x  2x
2sin
cos
+ 2cos
cos
=0
2
2
2
2
2sin 3x cos x + 2cos 3x cos x = 0
2cos x(sin 3x + cos 3x) = 0
cos x = 0
or
sin 3x = cos 3x

tan 3x = 1


Basic angle,  =
Basic angle,  =
2
4




x=
3x =   , 2  , 3 
2
4
4
4
  7 11
x = , ,
,
2 4 12 12
Ex 13: Further Trigonometric Identities
Prepared by Mr Ang KS
8
5.
6.
cos 6  cos 2
= tan 4 tan 2
cos 6  cos 2
6  2
6  2
6  2
6  2
L.H.S. = [2sin
sin
]  [2cos
cos
]
2
2
2
2
 2 sin 4 sin 2
=
2 cos 4 cos 2
= tan 4 tan 2
= R.H.S.
(c)
Prove:
(d)
Prove:
(f)
Prove: cos2 A  cos2 B = sin (B  A) sin (B + A)
L.H.S. = cos2 A  cos2 B
= (cos A + cos B)(cos A  cos B)
A B
A B
A B
A B
= (2cos
cos
)(2sin
sin
)
2
2
2
2
A B
A B
A B
A B
= (2sin
cos
)(2sin
cos
)
2
2
2
2
A B
A B
= sin 2(
) sin 2(
)
2
2
= sin [(B  A)] sin (A + B)
= sin (B  A) sin (B + A)
= R.H.S.
sin 6  cos 4  sin 2
= cot 4
cos 6  sin 4  cos 2
sin 6  sin 2  cos 4
L.H.S. =
cos 6  cos 2  sin 4
6  2
6  2
6  2
6  2
= [2cos
sin
+ cos 4]  [2sin
sin
 sin 4]
2
2
2
2
= [2cos 4 sin 2 + cos 4]  [2sin 4 sin 2  sin 4]
cos 4 (2 sin 2  1)
=
 sin 4 (2 sin 2  1)
= cot 4
= R.H.S.
1
4 x  3x
4 x  3x
[2sin
cos
]
2
2
2
7x
x
= sin
cos
2
2
7
1
1
a = 1, b =
= 3 and c = .
2
2
2
1
1
5x  4x
5x  4x
(sin 5x  sin 4x)
= [2cos
sin
]
2
2
2
2
9x
x
= cos
sin
2
2
1
(sin 4x + sin 3x)
2
=
Ex 13: Further Trigonometric Identities
Prepared by Mr Ang KS
9
7x
x
9x
x
cos
+ cos
sin = sin 4x cos x
2
2
2
2
1
1
L.H.S. = (sin 4x + sin 3x) + (sin 5x  sin 4x)
2
2
1
= (sin 5x + sin 3x)
2
1
5 x  3x
5 x  3x
= [2sin
cos
]
2
2
2
= sin 4x cos x
= R.H.S.
Show: sin
 End of Ex 13.4 
Ex 13: Further Trigonometric Identities
Prepared by Mr Ang KS
10