How to evaluate this sum
$\sum_{i=0}^{14}2^i\tan(2^i\theta) + 2^{15}\cot(2^{15}\theta)$
I need some hints on how to start
Evaluate $\sum_{i=0}^{14}2^i\tan(2^i\theta) + 2^{15}\cot(2^{15}\theta)$
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$\begingroup$
trigonometry
summation
2 Answers
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Hint: It follows from the identity $$\cot x-2\cot 2x =\tan x $$ This can be proved easily by using the identities of $\sin 2x $ and $\cos 2x $. Now we have, $$\cot \alpha -2\cot 2\alpha = \tan \alpha \tag {1} $$ $$2\cot 2\alpha -4\cot 4\alpha =2\tan 2\alpha \tag {2} $$ $$4\cot 4\alpha -8\cot 8\alpha =4\tan 4\alpha \tag {3} $$ and so on. Can you see a pattern visible now by adding them??
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HINT:
$$\cot x-\tan x=\dfrac{\cos^2x-\sin^2x}{\sin x\cos x}=2\cot2x$$
Replace $x$ with $\theta,2\theta,2^2\theta,\cdots,2^{15}\theta$