Question

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

Answer 1

`(sqrt6/4)(sqrt(2 - sqrt3))`

Explanation:

`f(x) = N/D = (sin^2 15 + sin 75)/(cos^2 36 + cos^2 54)`
First, evaluate the denominator D. Since,
`cos (54) = cos (90 - 54) = 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(sin 15 + cos 15) = sin 15(sqrt2cos (15 - 45))`
`N = sin 15(sqrt2cos (-30)) = (sqrt2)(sqrt3/2)sin 15 =`
`N = (sqrt6/2)sin 15`.
Find sin 15 by using trig identity
`2sin^2 a = 1 - cos 2a`. In this case -->
`2sin^2 15 = 1 - cos 30 = 1 - sqrt3/2 = (2 - sqrt3)/2`
`sin^2 15 = (2 - sqrt3)/4`
`sin 15 = sqrt(2 - sqrt3)/2` (since sin 15 is positive)
Finally
`f(x) = N/D = (sqrt6/4)(sqrt(2 - sqrt3))`