Page 49 - mathsvol1ch1to3ans

49

π θ 8
(ii) cos + , when sin θ = .
4 2 9
Solution:

π θ π θ π θ 1 θ θ
cos + = cos cos − sin sin = √ cos − sin
4 2 4 2 4 2 2 2 2
s
2
1 θ θ
= √ cos − sin
2 2 2
1 p
= √ (1 − sin θ)
2
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!
r
1 1
= √
2 9
1
= √
3 2

1 1 1 1
3
3. If cos θ = a + , show that cos 3θ = a + .
2 a 2 a 3

1 1
cos θ = a +
2 a
3
cos 3θ = 4 cos θ − 3 cos θ
2
= cos θ (4 cos θ − 3)
!
2
1 1 1
= a + a + − 3
2 a a
!
2
1 1 1
2
= a + a + − 1
2 a a

1 1
3
= a +
2 a 3
3
5
4. Prove that cos 5θ = 16 cos θ − 20 cos θ + 5 cos θ.
Solution:
cos 5θ = cos(3θ + 2θ)
= (cos 3θ cos 2θ − sin 3θ sin 2θ)
3
2
cos 3θ cos 2θ = (4 cos θ − 3 cos θ) (2 cos θ − 1)
5
3
= (8 cos θ − 10 cos θ + 3 cos θ)

3
sin 3θ sin 2θ = 3 sin θ − 4 sin θ (2 sin θ cos θ)
2
4
= 6 sin θ cos θ − 8 sin θ cos θ
4
2
3
sin 3θ sin 2θ = (6 cos θ − 6 cos θ − 8 (1 + cos θ − 2 cos θ) cos θ)
3
3
5
= (6 cos θ − 6 cos θ − 8 cos θ − 8 cos θ + 16 cos θ)