# How do you evaluate \frac { \sin ^ { 2} 15^ { \circ } + \sin 75^ { \circ } } { \cos ^ { 2} 36^ { \circ } + \cos ^ { 2} 54^ { \circ } }?

May 6, 2018

$\left(\frac{\sqrt{6}}{4}\right) \left(\sqrt{2 - \sqrt{3}}\right)$

#### Explanation:

$f \left(x\right) = \frac{N}{D} = \frac{{\sin}^{2} 15 + \sin 75}{{\cos}^{2} 36 + {\cos}^{2} 54}$
First, evaluate the denominator D. Since,
$\cos \left(54\right) = \cos \left(90 - 54\right) = \sin 36$
${\cos}^{2} 54 = {\sin}^{2} 36$ --> Therefor:
$D = {\cos}^{2} 36 + {\sin}^{2} 36 = 1$
Next, evaluate the numerator N.
Since sin 75 = cos (90 - 75) = cos 15. Therefor,-->
$N = \sin 15 \left(\sin 15 + \cos 15\right) = \sin 15 \left(\sqrt{2} \cos \left(15 - 45\right)\right)$
$N = \sin 15 \left(\sqrt{2} \cos \left(- 30\right)\right) = \left(\sqrt{2}\right) \left(\frac{\sqrt{3}}{2}\right) \sin 15 =$
$N = \left(\frac{\sqrt{6}}{2}\right) \sin 15$.
Find sin 15 by using trig identity
$2 {\sin}^{2} a = 1 - \cos 2 a$. In this case -->
$2 {\sin}^{2} 15 = 1 - \cos 30 = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$
${\sin}^{2} 15 = \frac{2 - \sqrt{3}}{4}$
$\sin 15 = \frac{\sqrt{2 - \sqrt{3}}}{2}$ (since sin 15 is positive)
Finally
$f \left(x\right) = \frac{N}{D} = \left(\frac{\sqrt{6}}{4}\right) \left(\sqrt{2 - \sqrt{3}}\right)$