# How do you find the exact functional value tan 165 using the cosine sum or difference identity?

Sep 21, 2015

$\frac{\sqrt{2 - \sqrt{3}}}{-} \sqrt{2 + \sqrt{3}}$

#### Explanation:

$\sin \left(165\right) = \sin \left(180 - 15\right) = \sin \left(15\right) = \sqrt{\frac{1 - \cos 2 \left(15\right)}{2}}$=$\sqrt{\frac{1 - \cos 30}{2}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \frac{\sqrt{2 - \sqrt{3}}}{2}$
cos(165)=cos(180-15)=-cos(15)=-sqrt((1+cos2(15))/2=$- \sqrt{\frac{1 + \cos 30}{2}} = - \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = - \frac{\sqrt{2 + \sqrt{3}}}{2}$
$\tan \left(165\right) = \sin \frac{165}{\cos} \left(165\right) = \frac{\sqrt{2 - \sqrt{3}}}{2} / - \frac{\sqrt{2 + \sqrt{3}}}{2} = \frac{\sqrt{2 - \sqrt{3}}}{-} \sqrt{2 + \sqrt{3}}$

Sep 21, 2015

Find tan (165)

Ans: $\frac{1 - \sqrt{3}}{1 + \sqrt{3}}$

#### Explanation:

Apply the trig identity: $\tan \left(a + b\right) = \frac{\tan a + \tan b}{1 - \tan a . \tan b}$
tan (165) = tan (45 + 120).
Trig table gives --> tan 45 = 1; $\tan 120 = - \sqrt{3}$

$\tan \left(165\right) = \frac{1 - \sqrt{3}}{1 + \sqrt{3}}$

Check by calculator: tan (165) = - 0.27
$\frac{1 - \sqrt{3}}{1 + \sqrt{3}} = - \frac{0.73}{2.73} = - 0.27$. OK