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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7 Answers
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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}
answered Jul 25, 2012 at 17:42
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$\begingroup$ $$\sin^2 A \cos^2 B -\sin^2 B \cos^2 A=\sin^2 A(1- \sin^2 B) -\sin^2 B (1-\sin^2 A)=?$$ $\endgroup$ Commented Aug 28, 2015 at 9:19
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$$ \begin{align*}\sin (A+B)\cdot\sin (A-B)&=\frac{\cos(2B)-\cos (2A)}{2}\\&=\frac{(1-2\sin^2B)-(1-2\sin^2A)}{2}\\&=\sin^2A-\sin^2B\end{align*}$$
Here i have used $$\sin x\cdot\sin y=\frac{\cos(x-y)-\cos(x+y)}{2}$$
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Hint: $(a+b)(a-b)=a^2-b^2$.
Then use $\sin^2\theta+\cos^2\theta=1$
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$\begingroup$ Do you mean I would do (sinA-sinB)(SinA+sinB)? $\endgroup$ Commented Jul 25, 2012 at 17:38
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$\begingroup$ No, $a=\sin A \cos B$ and $b=\sin B \cos A$ $\endgroup$– SaurabhCommented Jul 25, 2012 at 17:40
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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.
answered May 9, 2013 at 20:16
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A proof based on complex representation of sine: $$\begin{align} \sin(A+B)\sin(A-B)&=\frac{e^{i(A+B)}-e^{-i(A+B)}}{2i}\cdot\frac{e^{i(A-B)}-e^{-i(A-B)}}{2i}\\ &=\frac{e^{i2A}-e^{-i2B}-e^{i2B}+e^{-i2A}}{(2i)^2}\\ &=\left(\frac{e^{iA}-e^{-iA}}{2i}\right)^2 -\left(\frac{e^{iB}-e^{-iB}}{2i}\right)^2\\ &=\sin^2A-\sin^2B \end{align} $$
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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 :)
answered Feb 2, 2021 at 10:39
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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.
