# Prove $\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B$

How would I verify the following double angle identity. $$\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B$$ So far I have done this. $$(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)$$But I am not sure how to proceed.

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use this formula

$$2 \sin(A+B)\sin(A-B)=\cos2B-\cos2A$$

it will like this

$$\dfrac12 \cdot (\cos2B-\cos2A)$$ $$\dfrac {(1-2\sin^2B)-(1-2\sin^2A)}{2}$$

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\begin{eqnarray} \sin(A+B)\sin(A-B) &=& (\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)\\\\ &=& \sin^2 A \cos^2 B -\sin^2 B \cos^2 A\\\\ &=& \sin^2 A \cos^2 B -\sin^2 B (1-\sin^2 A)\\\\ &=& \sin^2 A (\cos^2 B + \sin^2 B) - \sin^2 B\\\\ &=& \sin^2 A - \sin^2 B \end{eqnarray}

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It makes sense.Thank you. –  El Bananero Jul 25 '12 at 17:47

\begin{align*}\sin (A+B)\cdot\sin (A-B)&=\frac{\cos(2B)-\cos (2A)}{2}\\&=\frac{(1-2\sin^2B)-(1-2\sin^2A)}{2}\\&=\sin^2A-\sin^2B\end{align*}

Here i have used $$\sin x\cdot\sin y=\frac{\cos(x-y)-\cos(x+y)}{2}$$

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Your question involves the basic algebra identity which says, $(a + b)(a - b) = a^2 - b^2$. For targeting your question, it is easy to assume $a = \sin A\cos B$ and $b = \cos A \sin B$. The process becomes easy now.

\begin{align}(a + b)(a - b)& =& a^2 - b^2\\ & = & (\sin A \cos B)^2 - (\cos A \sin B)^2\\ & = & \sin^2A\cos^2 B - \cos^2A\sin^2B \\ & = & \sin^2A(1 - \sin^2 B) - \cos^2 A\sin^2B \end{align}Proceed.

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Open the bracket of what you got and substitute every cos^2 with 1-sin^2 and open bracket again. Thats it...

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Hint: $(a+b)(a-b)=a^2-b^2$.
Then use $\sin^2\theta+\cos^2\theta=1$

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Do you mean I would do (sinA-sinB)(SinA+sinB)? –  El Bananero Jul 25 '12 at 17:38
No, $a=\sin A \cos B$ and $b=\sin B \cos A$ –  Saurabh Jul 25 '12 at 17:40
Oh I see thanks. –  El Bananero Jul 25 '12 at 17:47