Page 42 - mathsvol1ch1to3ans

42
√ √
1 3 1 − 3
◦
(i) cos 105 = cos(60 + 45) = cos(60)cos(45) − sin(60) sin(45) = √ − √ = √
2 2 2 2 √ 2 2
√
3 1 3 + 1
(ii) sin 105 ◦ = sin(60 + 45) = sin(60)cos(45) + cos(60) sin(45) = √ + √ = √
2 2 2 2 2 2
π π

7π π π tan + tan
(iii) tan = tan + = 4 3
12 4 3 1 − tan π tan π
√ 4 3
1 + 3 √
= √ = −(2 + 3)
1 − 3
√
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3 cos x − sin x
◦
6. Prove that (i) cos(30 + x) = (ii) cos(π + θ) = − cos θ
2
(iii) sin(π + θ) = − sin θ.
Solution:
√ √
3 sin x 3 cos x − sin x
◦
◦
◦
cos(30 + x) = cos(30 ) cos x − sin(30 ) sin x = cos x − =
2 2 2
cos(π + θ) = cos π cos θ − sin π sin θ = − cos θ
sin(π + θ) = sin π cos θ + cos π sin θ = − sin θ
◦
◦
7. Find a quadratic equation whose roots are sin 15 and cos 15 . r
√
√
3 − 1 3 + 1 3
◦
◦
Solution:We know that sin 15 = √ and cos 15 = √ . Now sum of the roots is =
2 2 2 2 2
1 √
2
and product of the roots = √ . The quadratic equation becomes 4x − 2 6x + 1 = 0.
2
8. Expand cos(A + B + C). Hence prove that
cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B,
π
if A + B + C = .
2
Solution:
cos(A + B + C) = cos[A + (B + C)]
= cos(A) cos(B + C) − sin(A) sin(B + C)

cos(A) cos(B + C) = cos(A)[cos(B) cos(C) − sin(B) sin(C)]

= cos(A) cos(B) cos(C) − cos(A) sin(B) sin(C)

sin(A) sin(B + C) = sin(A)[sin(B) cos(C) + sin(C) cos(B)]
= sin(A) sin(B) cos(C) + sin(A) sin(C) cos(B)

cos(A + B + C) = cos(A) cos(B) cos(C) − cos(A) sin(B) sin(C)

− sin(A) sin(B) cos(C) − sin(A) sin(C) cos(B)

π

Since cos = 0, cos(A + B + C) = 0. Hence we have
2
cos(A) cos(B) cos(C) = cos(A) sin(B) sin(C) + sin(A) sin(B) cos(C) + sin(A) sin(C) cos(B)
9. Prove that
√
◦
◦
(i) sin(45 + θ) − sin(45 − θ) = 2 sin θ.