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If the sides of a right-angled triangle are $\cos 2a + \cos 2b + 2\cos(a+b)$ and $\sin 2a + \sin 2b + 2\sin(a+b)$, find the hypotenuse- I can simplify this but I end up having a lot of terms in the end :/

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  • $\begingroup$ A shame ... This question was closed while I was off constructing a trigonographic answer. I've posted the solution to trigonography.com, anyway. $\endgroup$ – Blue Jun 4 '16 at 11:13
  • $\begingroup$ This question is of trigonometry -_- $\endgroup$ – Ankit Paul Jun 4 '16 at 14:47
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$$\cos 2a+\cos 2b+2\cos(a+b)=\cos^2 a-\sin^2 a + \cos^2b-\sin^2b+2\cos a\cos b-2\sin a\sin b=(\cos a+\cos b)^2-(\sin a+\sin b)^2$$ $$\sin2a+\sin2b+2\sin(a+b)=2\sin a\cos a+2\sin b\cos b+2\sin a\cos b+2\cos a\sin b=2(\sin a+\sin b)(\cos a+\cos b)$$ Let $\cos a+\cos b=u$, $\sin a+\sin b=v$. $$\text{Hypotenuse}=\sqrt{(u^2-v^2)^2+(2uv)^2}=u^2+v^2=(\cos a+\cos b)^2+(\sin a+\sin b)^2=2(1+\cos(a-b))$$

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HINT:

Use Prosthaphaeresis Formulas

$$\cos2A+\cos2B+2\cos(A+B)=2\cos(A+B)[1+\cos(A-B)]$$ $$=4\cos(A+B)\cdot\sin^2\dfrac{A-B}2$$

Similarly, $$\sin2A+\sin2B+2\sin(A+B)=\cdots=4\sin(A+B)\cdot\sin^2\dfrac{A-B}2$$

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Use Pythagorean theorem .

And use the identities $\cos^2 2a +sin ^2 2a =1$ , $\cos^2 2b +sin ^2 2b =1$ and $\cos^2(a+b) +\sin^2 (a+b)=1 $ to simplify the answer.

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