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Question: If 27 cos<sup>3</sup>q sin<sup>5</sup>q = a sin 8q + b sin 6q + c cos 4q + d sin 2q then value of a ...

If 27 cos3q sin5q = a sin 8q + b sin 6q + c cos 4q + d sin 2q then value of a + b + c + d is-

A

5

B

–6

C

0

D

3

Answer

3

Explanation

Solution

Q (2 cos q)3 = (z+1z)3\left( z + \frac{1}{z} \right)^{3}and

(2i sin q)5 = (z−1z)5\left( z - \frac{1}{z} \right)^{5}

Ž 28 i cos3 q sin5 q = (z2−1z2)3\left( z^{2} - \frac{1}{z^{2}} \right)^{3} (z−1z)2\left( z - \frac{1}{z} \right)^{2}

= (z6−3z2+31z2−1z6)\left( z^{6} - 3z^{2} + 3\frac{1}{z^{2}} - \frac{1}{z^{6}} \right) (z2−2+1z2)\left( z^{2} - 2 + \frac{1}{z^{2}} \right)

= (z8−1z8)\left( z^{8} - \frac{1}{z^{8}} \right)– 2 (z6−1z6)\left( z^{6} - \frac{1}{z^{6}} \right) – (3 – 1)

(z4−1z4)\left( z^{4} - \frac{1}{z^{4}} \right)+6(z2−1z2)6\left( z^{2} - \frac{1}{z^{2}} \right)

= (2i sin 8 q) – 2(2i sin 6q) – 2(2i sin 4q) + 6(2i sin 2q)

Ž 27 cos3q sin5q = sin 8q – 2sin 6q – 2sin 4q + 6 sin 2q

Here a = 1, b = –2, c = –2, d = 6

Ž a + b + c + d = 1 – 2 – 2 + 6 = 3.