# 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.

\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}

• It makes sense.Thank you. Commented Jul 25, 2012 at 17:47
• $$\sin^2 A \cos^2 B -\sin^2 B \cos^2 A=\sin^2 A(1- \sin^2 B) -\sin^2 B (1-\sin^2 A)=?$$ Commented Aug 28, 2015 at 9:19

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

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

• Do you mean I would do (sinA-sinB)(SinA+sinB)? Commented Jul 25, 2012 at 17:38
• No, $a=\sin A \cos B$ and $b=\sin B \cos A$ Commented Jul 25, 2012 at 17:40
• Oh I see thanks. Commented Jul 25, 2012 at 17:47

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 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}

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 :)

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.