I really don't know where I'm going wrong, I use the sum to product formula but always end up far from $\sin(2x)\sin(2y)$. Any help is appreciated, thanks.
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$\begingroup$ I think the way I've done it below is a bit more straightforward than the half-angle and power-reduction formulas proposed in a couple of others, but maybe everyone has to judge that for himself. $\endgroup$– Michael HardyCommented Mar 19, 2013 at 16:56
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$\begingroup$ Michael's seemed most intuitive to me, but I thank everyone for their help! $\endgroup$– seekerCommented Mar 19, 2013 at 18:23
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7 Answers
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\begin{align} \Big(\sin(x+y)\Big)^2 - \Big(\sin(x-y)\Big)^2 & = \Big( \sin x\cos y+\cos x\sin y \Big)^2 - \Big( \sin x\cos y+\cos x\sin y \Big)^2 \\[6pt] & = \Big( \sin^2\cos^2y + 2\sin x\cos y\cos x\sin y + \cos^2x\sin^2y \Big) \\[6pt] &\phantom{{}=} {}- \Big( \sin^2\cos^2y - 2\sin x\cos y\cos x\sin y + \cos^2x\sin^2y \Big) \\[6pt] & = 4\sin x\cos y\cos x\sin y \\[6pt] & = (2\sin x\cos x)(2\sin y\cos y) \\[6pt] & = \sin(2x)\sin(2y). \end{align}
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Known Identity
$$\sin^2 a = \frac{1-\cos 2a}{2}$$ $$\cos x - \cos y = -2 \sin( \frac{x - y}{2} )\sin( \frac{x + y}{2} ) $$
Given $$\sin^2(x+y)-\sin^2(x-y)$$ $$=\frac{1-\cos 2(x+y) -(1-\cos 2(x-y))}{2}$$ $$=\frac{\cos 2(x-y) - \cos 2(x+y)}{2}$$ $$=\sin(2x)\sin(2y)$$
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The following power-reduction formula should get you the rest of the way:
$$\sin^2\theta=\frac{1-\cos 2\theta}2$$
answered Mar 19, 2013 at 16:50
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Or you can use the very well-known identity: $$A^2-B^2=(A-B)(A+B)$$
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Just expand, note that $(a+b)^2-(a-b)^2 = 4 ab$.
Expand $\sin (x+y), \sin (x-y)$ in the usual way. Let $a = \sin x \cos y, b= \cos x \sin y$.
Then $\sin^2(x+y)−\sin^2(x−y)= 4 \sin x \cos y \cos x \sin y$.
Then note that $\sin 2 t = 2 \sin t \cos t$ to finish.
Note that the only trigonometric identity used here is $\sin (s+t) = \sin s \cos t + \sin t \cos s$.
answered Mar 19, 2013 at 16:57
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Apply $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$
Or
$\sin^2(x+y)-\sin^2(x-y)=\{\sin(x+y)+\sin(x-y)\}\{\sin(x+y)-\sin(x-y)\}$
Now (i)apply $\sin C+\sin D=2\sin\left(\frac {C+D}2\right)\cos\left(\frac {C-D}2\right)$ and $\sin C-\sin D=2\sin\left(\frac {C-D}2\right)\cos\left(\frac {C+D}2\right)$
Or (ii)use $\sin(A\pm B)=\sin A\cos B\pm \cos A\sin B$
answered Mar 19, 2013 at 17:11
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According to this:
$\sin(x+y) = \sin(x)\cos(y)+\cos(x)\sin(y)$
$\sin(x-y) = \sin(x)\cos(y)-\cos(x)\sin(y)$
$\sin^2(x+y) = {\sin^2(x)}{\cos^2(y)}+2\space{\sin(x)}{\cos(y)}{\cos(x)}{\sin(y)}+{\cos^2(x)}{\sin^2(y)} $, and
$\sin^2(x-y) = {\sin^2(x)}{\cos^2(y)}-2\space{\sin(x)}{\cos(y)}{\cos(x)}{\sin(y)}+{\cos^2(x)}{\sin^2(y)} $
${\sin^2(x+y)}-{\sin^2(x-y)}={4}\space{\sin(x)}{\cos(y)}{\cos(x)}{\sin(y)}=({2}\space{\sin(x)}{\cos(x)})\cdot({2}\space{\sin(y)}{\cos(y)})$
According to: ${2}\space\sin(\alpha)\cos(\alpha) = \sin({2}\alpha)$, you can write:
$${\sin^2(x+y)}-{\sin^2(x-y)} = \sin(2x)\cdot\sin(2y) $$.
It's all.
