Uploaded by Aryan Sharma

# alevelsb p1 ex10c

```Trigonometric identities and equations 10C
1
1
1 a As sin 2 θ + cos 2 θ ≡ 1
2
2
1
1
So 1 − cos 2 θ =
sin 2 θ
2
2
1 i
2
=1
5sin 3θ + 5cos 3θ = 5 ( sin 3θ + cos 3θ )
2
2
2
Given 2sin θ = 3cos θ
sin θ 3
So
=
cos θ 2
(divide both side by 2 cos θ )
3
So tan θ =
2
3
As sin x cos y = 3cos x sin y
sin x cos y
cos x sin y
So
=3
cos x cos y
cos x cos y
So tan x = 3 tan y
2
=5
c As sin 2 A + cos 2 A ≡ 1
So sin 2 A − 1 ≡ − cos 2 A
sin θ
sin θ
d
=
tan θ sin θ
cos θ
cos θ
= sin θ &times;
sin θ
= cos θ
g
θ + cos 2 θ )
2
2
f
2
= 12
So:
e
( sin
=
b As sin 3θ + cos 3θ ≡ 1
2
sin 4 θ + 2sin 2 θ cos 2 + cos 4 θ
4 a As sin 2 θ + cos 2 θ ≡ 1
So cos 2 θ ≡ 1 − sin 2 θ
1 − cos 2 x
sin 2 x
=
cos x
cos x
sin x
=
cos x
= tan x
1 − cos 2 3 A
1 − sin 2 3 A
=
b tan 2 θ ≡
c cos θ tan θ = cos θ &times;
= sin θ
sin 2 3 A
cos 2 3 A
sin 3 A
=
cos 3 A
= tan 3 A
(1 + sin x ) + (1 − sin x )
2
sin 2 θ
sin 2 θ
≡
cos 2 θ 1 − sin 2 θ
2
+ 2 cos 2 x
= 1 + 2sin x + sin 2 x + 1 − 2sin x
+ sin 2 x + 2 cos 2 x
=
2 + 2sin 2 x + 2 cos 2 x
=
2 + 2 ( sin 2 x + cos 2 x )
= 2+2
=4
h sin 4 θ + sin 2 θ cos 2 θ
= sin 2 θ ( sin 2 θ + cos 2 θ )
sin θ
cos θ
cos θ
cos θ
=
tan θ sin θ
cos θ
cos θ
= cos θ &times;
sin θ
2
cos θ
=
sin θ
cos θ 1 − sin 2 θ
1
=
− sin θ
So
or
tan θ
sin θ
sin θ
d
e
( cos θ − sin θ )( cos θ + sin θ )
= cos 2 θ − sin 2 θ
=
(1 − sin 2 θ ) − sin 2 θ
= 1 − 2sin 2 θ
= sin 2 θ
&copy; Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
1
5 a LHS
=
( sin θ + cos θ )
2
5 f
LHS =
2 − ( sin θ − cos θ )
2
=
2 − ( sin 2 θ − 2sin θ cos θ + cos 2 θ )
=
sin 2 θ + 2sin θ cos θ + cos 2 θ
= ( sin 2 θ + cos 2 θ ) + 2sin θ cos θ
= 2 − (1 − 2sin θ cos θ )
= 1 + 2sin θ cos θ
= 1 + 2sin θ cos θ
= RHS
= sin 2 θ + cos 2 θ + 2sin θ cos θ
1
b LHS
=
− cos θ
cos θ
1 − cos 2 θ
=
cos θ
sin 2 θ
=
cos θ
sin θ
= sin θ &times;
cos θ
= sin θ tan θ
=
1
tan x
sin x cos x
=
+
cos x sin x
sin 2 x + cos 2 x
=
sin x cos x
1
=
sin x cos x
= RHS
d LHS
= cos 2 A − sin 2 A
≡ cos 2 A − (1 − cos 2 A )
≡ cos 2 A − 1 + cos 2 A
≡ 2 cos 2 A − 1
≡ 2 (1 − sin 2 A ) − 1
2
= RHS
g LHS sin 2 x cos 2 y − cos 2 x sin 2 y
=
= sin 2 x (1 − sin 2 y )
− (1 − sin 2 x ) sin 2 y
= sin 2 x − sin 2 x sin 2 y
− sin 2 y + sin 2 x sin 2 y
= sin 2 x − sin 2 y
= RHS
= RHS
c LHS
= tan x +
( sin θ + cos θ )
6 a
Using Pythagoras' theorem:
x 2 = 122 + 52 = 169
x = 13
5
12
So sin θ =
and cos θ
=
13
13
b
≡ 2 − 2sin 2 A − 1
≡ 1 − 2sin 2 A
e LHS = (2sin θ − cos θ ) 2 + (sin θ + 2 cos θ ) 2
≡ 4sin 2 θ − 4sin θ cos θ + cos 2 θ
+ sin 2 θ + 4sin θ cos θ + 4 cos 2 θ
≡ 5sin 2 θ + 5cos 2 θ
Using Pythagoras' theorem, x = 4
≡ 5 ( sin 2 θ + cos 2 θ )
So sin φ =
4
5
and tan φ = − 43
≡5
≡ RHS
&copy; Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
2
6 b
7
Consider the angle φ where sin φ = 23 .
Using Pythagoras' theorem, x = 5
a So cos φ =
As θ is obtuse:
sin
=
θ sin
=
φ
5
3
4
5
and
tan θ =
− tan φ =
− 43
c
5
As θ is obtuse, cos θ =
− cos φ =
−
3
Using Pythagoras’ theorem
x2 + 72 =
252
=
x 2 252 − 7 2
= 576
x = 24
=
So cos φ 24
=
and tan φ
25
b From the triangle
2
tan φ =
5
2 5
5
Using the quadrant diagram
tan θ = − tan φ
=
7
24
= −
8
2 5
5
Draw a right-angled triangle with
tan φ = + 3
=
3
1
As θ is in the fourth quadrant
cos θ = + cos φ
=
24
25
and tan θ = − tan φ
= − 247
Using Pythagoras’ theorem
2
x=
( 3)
2
+=
12 4
So x = 2
&copy; Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
3
8 a sin φ =
9 a sin θ = − sin φ
3
2
7
4
b tan θ = − tan φ
= −
= −
7
3
10 a As sin 2 θ + cos 2 θ ≡ 1
x2 + y 2 =
1
As θ is reflex and tan φ is –ve, φ is in the
So sin θ = − sin φ
2
− 3
=
2
b cos φ = 12
=
As cos θ cos
=
φ , cos θ
9
y
2
2
So, using sin θ + cos 2 θ ≡ 1
b sin θ x=
and cos θ
=
 y
x +  =
1
2
y2
or x 2 +
=
1
4
or 4 x 2 + y 2 =
4
2
1
2
Draw a right-angled triangle with
cos φ = 34 .
c As sin θ = x
sin 2 θ = x 2
Using sin 2 θ + cos 2 θ ≡ 1
x2 + y =
1
d As tan θ =
Using Pythagoras’ theorem
42
x 2 + 32 =
2
42 − 32
x=
=7
x= 7
7
=
So sinφ =
and tan φ
4
sin θ
cos θ
sin θ
tan θ
x
So cos θ =
y
cos θ =
Using sin 2 θ + cos 2 θ ≡ 1
x2
1 or x 2 y 2 =
x +=
+ x2 y 2
2
y
2
7
3
e sin θ + cos θ =
x
y
− sin θ + cos θ =
As θ is reflex and cos θ is +ve, θ is in
Adding the two equations:
2 cos θ= x + y
x+ y
So cos θ =
2
Subtracting the two equations:
2sin θ= x − y
x− y
So sin θ =
2
&copy; Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
4
10 e Using sin 2 θ + cos 2 θ ≡ 1
 x− y  x+ y
1

 +
 =
 2   2 
4
x 2 − 2 xy + y 2 + x 2 + 2 xy + y 2 =
2
2
2x2 + 2 y 2 =
4
2
x2 + y 2 =
11 a Using the cosine rule
a 2 + c2 − b2
cos B =
2ac
2
8 + 122 − 102
cos B =
2 &times; 8 &times;12
64 + 144 − 100
cos B =
192
108
cos B = 192
cos B = 169
b Since sin2θ + cos2θ = 1
2
sin2 B + ( 169 ) = 1
12 a Using the sine rule
sin Q sin P
=
q
p
sin Q sin 30&deg;
=
8
6
8sin 30&deg;
sin Q =
6
8 &times; 12
=
6
2
=
3
b Since sin2 θ + cos2 θ = 1
2
( 23 ) + cos2 Q = 1
cos2 Q = 1− 94
= 95
Since Q is obtuse Q is in the second
quadrant where cosine is negative.
5
So cos Q = −
3
81
sin2 B = 1− 256
= 175
256
So sin B =
=
175
256
5 7
16
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5
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