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Let $\alpha$ and $\beta$ be such that $\pi<\alpha-\beta<3\pi$, and $cos\alpha+cos\beta=-27/65$ and $sin\alpha+sin\beta= -21/65$, find $cos(\alpha-\beta)/2$

I solved it by squaring the two equations such that I get $sin^2\alpha+sin^2beta+2sin\alpha sin\beta=441/4225$ And $cos^2\alpha+cos^\beta+2cos\alpha cos\beta=729/4225$

Adding them we get $2+2(sin\alpha sin\beta + cos\alpha cos\beta)= 1170/4225$

Then, $1+cos(\alpha-\beta)=585/4225$

Therefore $cos(\alpha-\beta)=-3640/4225$

So now $cos(\alpha-\beta)/2=+or-\sqrt\frac{585}{8450}$

This is my answer. But the answer is -$\frac{3}{\sqrt130}$. What went wrong in my solving?

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    $\begingroup$ $$585 = 3^2\times5\times13$$ $$8450 = 2\times5^2\times13^2$$ So both are equivalent. $\endgroup$ – Ak19 Jul 30 '19 at 13:45
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$$\sqrt{\frac{585}{8450}} = \sqrt{\frac{65 \times 9}{8450}} = 3\sqrt{\frac{65 }{8450}} = 3\sqrt{\frac{1}{130}}$$ also your minus comes from $\alpha-\beta \in [\pi,3\pi]$

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  • $\begingroup$ Wait I got it, thanks! $\endgroup$ – Aditya Jul 30 '19 at 13:50
  • $\begingroup$ because $\alpha - \beta$ is in $[\pi,3\pi]$ $\endgroup$ – Ahmad Bazzi Jul 30 '19 at 13:50
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The answer is $-\frac{3}{\sqrt{130}}$ because $$\frac{\pi}{2}<\frac{\alpha-\beta}{2}<\frac{3\pi}{2},$$ which gives $$\cos\frac{\alpha-\beta}{2}<0.$$

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