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I've tried starting from both sides, the furthest I've gone is by starting from the right hand side: $$\begin{align*} \sin{11\theta}\sin{3\theta}&=\sin{(7\theta+4\theta)}\sin{(7\theta-4\theta)} \\ \\ &=(\sin{7\theta}\cos{4\theta}+\sin{4\theta}\cos{7\theta})(\sin{7\theta}\cos{4\theta}-\sin{4\theta}\cos{7\theta}) \\ \\ &=\sin^2{7\theta}\cos^2{4\theta}-\sin^2{4\theta}\cos^2{7\theta} \end{align*}$$

Have I done the identity wrong, or should I be using a different approach?

MathFail
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edds_7575
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  • In the equality you want to prove, expand each side using Euler's formula $\sin x=\frac{e^{ix}-e^{-ix}}{2i}$ and you get instantaneously the result. – Anne Bauval May 06 '23 at 15:56

2 Answers2

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$$\begin{align*} \sin{11\theta}\sin{3\theta}&=\sin{(7\theta+4\theta)}\sin{(7\theta-4\theta)} \\ \\ &=(\sin{7\theta}\cos{4\theta}+\sin{4\theta}\cos{7\theta})(\sin{7\theta}\cos{4\theta}-\sin{4\theta}\cos{7\theta}) \\ \\ &=\sin^2{7\theta}\cos^2{4\theta}-\sin^2{4\theta}\cos^2{7\theta} \end{align*}$$

You are almost there, keep going,

$$\begin{align}\sin^2{7\theta}\cos^2{4\theta}-\sin^2{4\theta}\cos^2{7\theta}&=\sin^2{7\theta}(1-\sin^2{4\theta})-\sin^2{4\theta}(1-\sin^2{7\theta})\\ \\ &=\sin^2{7\theta}-\sin^2{4\theta}\end{align}$$

MathFail
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2

$\sin^2{7\theta}-\sin^2{4\theta}=(\sin7\theta - \sin4\theta)\times (\sin7\theta + \sin4\theta)$

$2 \sin(\frac{7\theta - 4\theta}{2}) \cos(\frac{7\theta + 4\theta}{2}) \times 2 \sin(\frac{7\theta + 4\theta}{2}) \cos(\frac{7\theta - 4\theta}{2})=$

$= 2 \sin(\frac{3\theta}{2}) \cos(\frac{11\theta}{2}) \times 2 \sin(\frac{11\theta}{2}) \cos(\frac{3\theta}{2})=$

$=2 \sin(\frac{3\theta}{2})\cos(\frac{3\theta}{2})\times 2 \sin(\frac{11\theta}{2})\cos(\frac{11\theta}{2})=$

$=\sin3\theta \sin11\theta$

Lion Heart
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