# Thread: another trig identity question

1. ## another trig identity question

Hi
I need help the following:

Given $0 < \alpha,\gamma < \pi$ and $cos(\alpha) = \frac{2}{3}$, $cos(\gamma) = \frac{-1}{3}$ calculate exact expression for $cos(\alpha - \gamma)?$

This is what i have done however it is incorrect:

we know that $cos(\alpha) = \frac{2}{3}, cos(\gamma) = \frac{-1}{3},
sin(\alpha) = \frac{3}{\sqrt{13}} and sin(\gamma) = \frac{-26}{81}$

i sub it into: $cos(\alpha)cos(\gamma) + sin(\alpha)sin(\gamma)$

so i get: $\frac{-\sqrt{13}}{27} + \frac{-78}{81\sqrt{13}}$

P.S

2. Originally Posted by Paymemoney
Hi
I need help the following:

Given $0 < \alpha,\gamma < \pi$ and $cos(\alpha) = \frac{2}{3}$, $cos(\gamma) = \frac{-1}{3}$ calculate exact expression for $cos(\alpha - \gamma)?$

This is what i have done however it is incorrect:

we know that $cos(\alpha) = \frac{2}{3}, cos(\gamma) = \frac{-1}{3},
sin(\alpha) = \frac{3}{\sqrt{13}} and sin(\gamma) = \frac{-26}{81}$

i sub it into: $cos(\alpha)cos(\gamma) + sin(\alpha)sin(\gamma)$

so i get: $\frac{-\sqrt{13}}{27} + \frac{-78}{81\sqrt{13}}$

P.S
Dear Paymemoney,

$cos\alpha = \frac{2}{3}$

$sin\alpha=\sqrt{1-cos^2\alpha}=\frac{\sqrt{5}}{3}$

$cos\gamma = \frac{-1}{3}$

$sin\gamma=\sqrt{1-cos^2\gamma}=\frac{\sqrt{8}}{3}$

Can you continue from here??

3. yep i got the answer