Question

If sin 5theta = a sin^5+b sin^3theta+csin theta+d, then​

Answer 1

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Answer 2

Answer:

The actual question must be $sin 5\theta = a sin^{5}\theta + b sin^{3}\theta + c sin \theta + d$, then find a,b,c & d.

Step-by-step explanation:

We can solve this by using

sin(A+B) = sinAcosB + cosAsinB.

$sin 5\theta = sin(3\theta+2\theta)\\ sin(3\theta+2\theta) = sin3\theta cos2\theta + cos3\theta sin2\theta\\ sin(3\theta+2\theta) = (3sin\theta-4sin^{3}\theta)(1-2sin^2{\theta}) + (4cos^3{\theta}-3cos\theta)(2sin\theta cos\theta)$

Here we have used expansion formulae of sin 3A, cos 3A and sin 2A.

$sin(3\theta + 2\theta) = 3sin\theta - 6sin^3{\theta} - 4sin^3{\theta} + 8sin^5{\theta} + 8 sin\theta cos^4{\theta} - 6 sin\theta cos^2{\theta}\\sin(3\theta+2\theta) = 3sin\theta -10sin^3{\theta} + 8sin^5{\theta} + 8 sin\theta(1-sin^2{\theta})^2 - 6sin\theta (1 - sin^2{\theta})\\sin(3\theta+2\theta) = 3sin\theta -10sin^3{\theta} + 8sin^5{\theta} + 8 sin\theta(1 + sin^4{\theta - 2sin^2{\theta}) - 6sin\theta + 6 sin^3{\theta}$

$sin(3\theta+2\theta) = 16sin^5\theta - 20sin^3{\theta} +5 sin\theta$

Hence on comparing LHS with RHS we get,

a = 16, b = -20, c = 5 and d = 0.