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I tried 5 times proving the above identity but I ended up in the same way as shown below (Sorry for the bad handwriting)

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I tried substituting $n=0$ to find out whether the question is correct in the first place and after substituting n with 0 I got

$\sin^2 A = \sin^2A - \sin^2 (-A) = 0$

Since the above equation would only be true if $A=0$ is the question correct in the first place

I have only started learning trigonometry like 2 months before and I'm not pretty sure about myself

Thank you in advance :)

sillysillybag69
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  • Yeah, it is wrong, the right side should be $\sin^2(n+1)A-\sin^2 nA,$ or the left side should be $\sin 2nA\cdot\sin2A.$ These follow from the fun identity: $$\sin(b+c)\sin(b-c)=\sin^2b-\sin^2c,\tag1$$ when $b=(n+1)A$ and $c=nA$ in the first case an $b=(n+1)A,c=(n-1)A$ for the second case. You'd probably have to prove $(1).$ – Thomas Andrews Sep 02 '23 at 14:02
  • Your final result is correct, and the proposed equation was false for every $n$ (not only for $n=0$), as you can see by Tayor approximation as $A\to0$. – Anne Bauval Sep 02 '23 at 14:02

1 Answers1

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There is a typo.

$$\text{lhs - rhs}=\frac{1}{2} (\cos (2 A n)-\cos (2 A (n-1)))$$

As @Anne Bauval suggested, using Taylor expansions $$\text{lhs - rhs}=A^2 (1-2 n)+\frac{1}{3} A^4 \left(4 n^3-6 n^2+4 n-1\right)+O\left(A^6\right)$$

Claude Leibovici
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