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.
 A: Hint: $(a+b)(a-b)=a^2-b^2$.
Then use $\sin^2\theta+\cos^2\theta=1$
A: Use this formula:
$$2 \sin(A+B)\sin(A-B)=\cos2B-\cos2A$$
It will be like this:
$$\dfrac12 \cdot (\cos2B-\cos2A)$$
$$=\dfrac {(1-2\sin^2B)-(1-2\sin^2A)}{2}$$
It will give the answer if you simplify.
A: \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}
A: A proof based on complex representation of sine:
$$\begin{align}
\sin(A+B)\sin(A-B)&=\frac{e^{i(A+B)}-e^{-i(A+B)}}{2i}\cdot\frac{e^{i(A-B)}-e^{-i(A-B)}}{2i}\\
&=\frac{e^{i2A}-e^{-i2B}-e^{i2B}+e^{-i2A}}{(2i)^2}\\
&=\left(\frac{e^{iA}-e^{-iA}}{2i}\right)^2
-\left(\frac{e^{iB}-e^{-iB}}{2i}\right)^2\\
&=\sin^2A-\sin^2B
\end{align}
$$
A: $$ \begin{align*}\sin (A+B)\cdot\sin (A-B)&=\frac{\cos(2B)-\cos (2A)}{2}\\&=\frac{(1-2\sin^2B)-(1-2\sin^2A)}{2}\\&=\sin^2A-\sin^2B\end{align*}$$
Here i have used $$\sin x\cdot\sin y=\frac{\cos(x-y)-\cos(x+y)}{2}$$
A: 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.
A: I will prove the result by starting with the right hand side of the identity:
$$\begin{align}\sin^2A-\sin^2B&=(\sin A+\sin B )(\sin A-\sin B)\\
&=(2\sin\frac{A+B}{2}\cos\frac{A-B}{2})(2\sin\frac{A-B}{2}\cos\frac{A+B}{2})\\
&=(2\sin\frac{A+B}{2}\cos\frac{A+B}{2})(2\sin\frac{A-B}{2}\cos\frac{A-B}{2})\\
&=\sin(A+B)\sin(A-B)
\end{align}$$
as required,  on using the sum to product formulae in the second line of working and the double angle formulae in  the final line. I hope you found that interesting :)
