Page 48 - mathsvol1ch1to3ans

48

Given that k tan φ = tan θ

tan θ
Hence k =
tan φ
sin θ sin φ
−
k − 1 tan θ − tan φ cos θ cos φ
Using Componendo & Dividendo rule = =
k + 1 tan θ + tan φ sin θ sin φ
+
cos θ cos φ
sin θ cos φ − cos θ sin φ
cos θ cos φ
=
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sin θ cos φ + cos θ sin φ
cos θ cos φ
sin (θ − φ)
=
sin (θ + φ)

sin (θ − φ)
=
sin α
Exercise 3.5:

1. Find the value of cos 2A, A lies in the first quadrant, when
15 4 16
(i) cos A = (ii) sin A = (iii) tan A = .
17 5 63
Solution:
15 161
2
2
(i) cos 2A = 2 cos A − 1 = 2 − 1 =
17 289
2
4 7
2
(ii)cos 2A = 1 − 2 sin A = 1 − 2 = −
5 25
256
2
1 − tan A 1 − 3713
(iii)cos 2A = = 3969 =
2
1 + tan A 256 4225
1 +
3969
2. If θ is an acute angle, then find

π θ 1
(i) sin − , when sin θ = .
4 2 25
Solution:

π θ π θ θ π 1 θ θ
sin − = sin cos − sin cos = √ cos − sin
4 2 4 2 2 4 2 2 2
s
2
1 θ θ
= √ cos − sin
2 2 2
1 p
= √ (1 − sin θ)
2
!
r
1 24
= √
2 25
√
2 3
=
5