# Find the value of the function $cos(\alpha-\beta)/2$

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?

• $$585 = 3^2\times5\times13$$ $$8450 = 2\times5^2\times13^2$$ So both are equivalent. – Ak19 Jul 30 '19 at 13:45

$$\sqrt{\frac{585}{8450}} = \sqrt{\frac{65 \times 9}{8450}} = 3\sqrt{\frac{65 }{8450}} = 3\sqrt{\frac{1}{130}}$$ also your minus comes from $$\alpha-\beta \in [\pi,3\pi]$$
• because $\alpha - \beta$ is in $[\pi,3\pi]$ – Ahmad Bazzi Jul 30 '19 at 13:50
The answer is $$-\frac{3}{\sqrt{130}}$$ because $$\frac{\pi}{2}<\frac{\alpha-\beta}{2}<\frac{3\pi}{2},$$ which gives $$\cos\frac{\alpha-\beta}{2}<0.$$