Page 50 - mathsvol1ch1to3ans

50

3
5
3
5
3
cos 5θ = (8 cos θ − 10 cos θ + 3 cos θ) − (6 cos θ − 6 cos θ − 8 cos θ − 8 cos θ + 16 cos θ)
5
3
= 16 cos θ − 20 cos θ + 5 cos θ
2
1 − tan α
5. Prove that sin 4α = 4 tan α .
2
(1 + tan α) 2
Solution:
sin 4α = 2 sin 2α cos 2α
2
2 tan α 1 − tan α
2 sin 2α cos 2α = 2 ×
2
2
1 + tan α 1 + tan α
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2
tan α (1 − tan α)
= 4 2
2
(1 + tan α)
◦
6. If A + B = 45 , show that (1 + tan A) (1 + tan B) = 2.
Solution:
tan A + tan B
tan(A + B) = = 1
1 − tan A tan B
tan A + tan B + tan A tan B = 1

1 + tan A + tan B + tan A tan B = 2

(1 + tan A) + tan B(1 + tan A) = 2

(1 + tan A)(1 + tan B) = 2

◦
◦
◦
◦
7. Prove that (1 + tan 1 )(1 + tan 2 )(1 + tan 3 ) . . . (1 + tan 44 ) is a multiple of 4.
Solution:
◦
If A + B = 45 ,then (1 + tan A) (1 + tan B) = 2.Here we can rearrange the terms as,
◦
◦
◦
◦
◦
(1 + tan 1 )(1 + tan 44 )(1 + tan 2 ) . . . (1 + tan 22 )(1 + tan 23 ) = 2(22 pairs) =
44( is a multiple of 4)
π π

8. Prove that tan + θ − tan − θ = 2 tan 2θ.
4 4
Solution:
π π 1 + tan θ 1 − tan θ

tan + θ − tan − θ = −
4 4 1 − tan θ 1 + tan θ
2 2
(1 + tan θ) − (1 − tan θ)
=
2
1 − tan θ
4 tan θ
=
2
1 − tan θ
= 2 tan 2θ
◦ √ √ √ √
1
9. Show that cot 7 = 2 + 3 + 4 + 6.
2
Solution: