Series problem involving sine function

Question

Prove that:

$$ \sum 2^{2n}\sin^4\left(a \over 2^{n}\right) = 4^n\sin^{2}\left(a \over 2^{n} \right) - \sin^{2}\left(a\right) $$

I tried converting the square power on sin by trying the half angle formula, I saw a relation between the 2 sides of equality, its like substituting n in place of n and taking square root of sin function minus putting 0 on LHS and taking square root of sine. Its just a thought.

score 3 · Accepted Answer · answered Jan 09 '18 at 18:11

3

$$4^m\sin^4\dfrac a{2^m}-4^m\sin^2\dfrac a{2^m}=4^m\sin^2\dfrac a{2^m}\left(\sin^2\dfrac a{2^m}-1\right)=-4^{m-1}\left(2\sin\dfrac a{2^m}\cos\dfrac a{2^m}\right)^2$$

$$=-4^{m-1}\sin^2\dfrac a{2^{m-1}}$$

$$\implies4^m\sin^4\dfrac a{2^m}=4^m\sin^2\dfrac a{2^m}-4^{m-1}\sin^2\dfrac a{2^{m-1}}$$

Recognize the Telescoping series

Set $m=1,2,\cdots,n-1,n$ and add

answered Jan 09 '18 at 18:11

lab bhattacharjee

274,582

yeah i recognised it in the final step :) – Shashank Shekhar Jan 09 '18 at 18:13
@ShashankShekhar, Do you have any confusion? – lab bhattacharjee Jan 09 '18 at 18:14
yeah..how did you think of the first step ? how did it click ? – Shashank Shekhar Jan 09 '18 at 18:15
@ShashankShekhar, Subtracted the first term of the right hand side from the final term of the left hand side – lab bhattacharjee Jan 09 '18 at 18:16
okay..that was a great approach ! – Shashank Shekhar Jan 09 '18 at 18:18
why did you do that..i mean how did you think of doing this ? – Shashank Shekhar Jan 09 '18 at 18:19
@ShashankShekhar, Like https://math.stackexchange.com/questions/1591220/frac1-sin-8-circ-frac1-sin-16-circ-frac1-sin-4096-circ, I perceived that Telescoping Series will be identified. – lab bhattacharjee Jan 09 '18 at 18:22

Series problem involving sine function

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