Trigonometry calculation problem

Question

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.

Answer 1

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

You are almost there!

Answer 2

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:

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