# Trigonometry calculation problem

Problem: Calculate $\cos(\alpha+60)$, if $\sin(\alpha)=\frac{2}{3}$ and $\alpha\in[90^\circ,180^\circ]$.

I have tried following:

$$\cos(60^\circ)\cos\alpha-\sin(60^\circ)\sin\alpha=\cos(\alpha+60^\circ)\\\frac{\cos\alpha}{2}-\frac{\sqrt{3}}{3}=\cos(\alpha+60^\circ)$$

I would appreciate any hints.

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Small mistake on $\sin60^o$ – Mike Jul 7 '14 at 22:19
yeah, thanks !!! – Invisible Jul 7 '14 at 22:19
$\TeX$ and its descendants such as MathJax were not designed by primitive cave men. Accordingly I changed $90^o$ to $90^\circ$. ${}\qquad{}$ – Michael Hardy Jul 8 '14 at 0:07

Hint: $\sin^2(\alpha)+\cos^2(\alpha)=1$

You are almost there!

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Hey, thank you for a hint but I don't see the connection between what you wrote and my problem, can you elaborate? – Invisible Jul 7 '14 at 22:21
@Heisenberg Using $\sin^2(\alpha)+\cos^2(\alpha)=1$ we can calculate $\cos(\alpha)$ because we already know $\sin(\alpha)$ (and don't forget that we know $\alpha\in[90^\circ,180^\circ]$, so $\cos(\alpha)$ will be negative). – Peter Woolfitt Jul 7 '14 at 22:25

Using the identity $$\sin^2 \alpha + \cos^2 \alpha = 1$$

$$\frac{4}{9} + \cos^2 \alpha = 1$$ $$\cos^2 \alpha = \frac{5}{9}$$ $$\cos \alpha = \frac{\sqrt{5}}{3}$$

Using another identity:

$$\cos(60^o)\cos\alpha-\sin(60^o)\sin\alpha=\cos(\alpha+60^o)\\\frac{\cos\alpha}{2}-\frac{\sqrt{3}}{3}=\cos(\alpha+60^o)$$ $$\cos(\alpha+60^o) = \frac{\frac{\sqrt{5}}{3}}{2} - \frac{\sqrt{3}}{3}$$ $$\cos(\alpha+60^o) = \frac{\sqrt{5}-2\sqrt{3}}{6}$$

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