Prove that: $\operatorname{cosec}(2A) + \operatorname{cosec}(4A) + \operatorname{cosec}(8A) = \cot(A) - \cot(8A) $

Question

How can we prove $\operatorname{cosec}(2A) + \operatorname{cosec}(4A) + \operatorname{cosec}(8A) = \cot(A) - \cot(8A)$?

Answer 1

Put the right side to the left. $$cosec(2A)+cosec(4A)+cosec(8A)-cot(A)+cot(8A)=0$$ Then express the left side by sines and cosines. You get: $$\frac{1}{sin(2A)}+\frac{1}{sin(4A)}+\frac{1}{sin(8A)}-\frac{cos(A)}{sin(A)}+\frac{cos(8A)}{sin(8A)} =$$ $$= \frac{1}{sin(2A)}+\frac{1}{sin(4A)}+\frac{1+cos(8A)}{sin(8A)}-\frac{cos(A)}{sin(A)} =$$ $$= \frac{1}{sin(2A)}+\frac{1}{sin(4A)}+\frac{2cos^2(4A)}{2sin(4A)cos(4A)}-\frac{cos(A)}{sin(A)} = $$ $$= \frac{1}{sin(2A)}+\frac{1}{sin(4A)}+\frac{cos(4A)}{sin(4A)}-\frac{cos(A)}{sin(A)} = $$ $$= \frac{1}{sin(2A)}+\frac{1+cos(4A)}{sin(4A)}-\frac{cos(A)}{sin(A)} = $$ $$= \frac{1}{sin(2A)}+\frac{2cos^2(2A)}{2sin(2A)cos(2A)}-\frac{cos(A)}{sin(A)} = $$ $$= \frac{1}{sin(2A)}+\frac{cos(2A)}{sin(2A)}-\frac{cos(A)}{sin(A)} = $$ $$= \frac{1+cos(2A)}{sin(2A)}-\frac{cos(A)}{sin(A)} = $$ $$= \frac{2cos^2(A)}{2sin(A)cos(A)}-\frac{cos(A)}{sin(A)} = $$ $$= \frac{cos(A)}{sin(A)}-\frac{cos(A)}{sin(A)} = 0$$

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