Find:$$\dfrac{\sin(A+B)\sin(A-B)}{\sin A\cos A-\sin B\cos B}$$
I couldn't simplify afterwards so help me out
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Michael Rozenberg
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user466377
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1Is $a$ different from $A$? Is $b$ different from $B$. Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Jul 23 '17 at 10:35
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Everything is in capital letters – user466377 Jul 23 '17 at 10:37
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Please share your thoughts so far. – Shaun Jul 23 '17 at 10:58
2 Answers
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HINT:
Use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $
then $$2\sin A\cos A=\sin2A$$
finally Prosthaphaeresis Formulas $$\sin C-\sin D=?$$
lab bhattacharjee
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The original problem was $$\frac{\sin^2a-\sin^2b}{\sin{a}\cos{a}-\sin{b}\cos{b}},$$ which is $$\frac{\frac{1-\cos2a}{2}-\frac{1-\cos2b}{2}}{\frac{1}{2}\sin2a-\frac{1}{2}\sin2b}$$ or $$\frac{\cos2b-\cos2a}{\sin2a-\sin2b}$$ or $$\frac{\sin(a-b)\sin(a+b)}{\sin(a-b)\cos(a+b)}$$ or $$\tan(a+b)$$
Michael Rozenberg
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Is that a direct formula or based on numerator and denominator or are there some to that? – user466377 Jul 23 '17 at 10:40
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@user466377 I solved you previous problem: $\frac{\sin^2a-\sin^2b}{\sin{a}\cos{a}-\sin{b}\cos{b}}$ – Michael Rozenberg Jul 23 '17 at 10:42