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

Question

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

1 Answer

Impact of this question

1931 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-06T14:46:06

$(\frac{\sqrt{6}}{4}) (\sqrt{2 - \sqrt{3}})$ $(sqrt6/4)(sqrt(2 - sqrt3))$

Explanation:

$f (x) = \frac{N}{D} = \frac{{sin}^{2} 15 + sin 75}{{cos}^{2} 36 + {cos}^{2} 54}$ $f(x) = N/D = (sin^2 15 + sin 75)/(cos^2 36 + cos^2 54)$
First, evaluate the denominator D. Since,
$cos (54) = cos (90 - 54) = sin 36$ $cos (54) = cos (90 - 54) = sin 36$
${cos}^{2} 54 = {sin}^{2} 36$ $cos^2 54 = sin^2 36$ --> Therefor:
$D = {cos}^{2} 36 + {sin}^{2} 36 = 1$ $D = cos^2 36 + sin^2 36 = 1$
Next, evaluate the numerator N.
Since sin 75 = cos (90 - 75) = cos 15. Therefor,-->
$N = sin 15 (sin 15 + cos 15) = sin 15 (\sqrt{2} cos (15 - 45))$ $N = sin 15(sin 15 + cos 15) = sin 15(sqrt2cos (15 - 45))$
$N = sin 15 (\sqrt{2} cos (- 30)) = (\sqrt{2}) (\frac{\sqrt{3}}{2}) sin 15 =$ $N = sin 15(sqrt2cos (-30)) = (sqrt2)(sqrt3/2)sin 15 =$
$N = (\frac{\sqrt{6}}{2}) sin 15$ $N = (sqrt6/2)sin 15$ .
Find sin 15 by using trig identity
$2 {sin}^{2} a = 1 - cos 2 a$ $2sin^2 a = 1 - cos 2a$ . In this case -->
$2 {sin}^{2} 15 = 1 - cos 30 = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$ $2sin^2 15 = 1 - cos 30 = 1 - sqrt3/2 = (2 - sqrt3)/2$
${sin}^{2} 15 = \frac{2 - \sqrt{3}}{4}$ $sin^2 15 = (2 - sqrt3)/4$
$sin 15 = \frac{\sqrt{2 - \sqrt{3}}}{2}$ $sin 15 = sqrt(2 - sqrt3)/2$ (since sin 15 is positive)
Finally
$f (x) = \frac{N}{D} = (\frac{\sqrt{6}}{4}) (\sqrt{2 - \sqrt{3}})$ $f(x) = N/D = (sqrt6/4)(sqrt(2 - sqrt3))$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you evaluate sin215∘+sin75∘cos236∘+cos254∘\frac { \sin ^ { 2} 15^ { \circ } + \sin 75^ { \circ } } { \cos ^ { 2} 36^ { \circ } + \cos ^ { 2} 54^ { \circ } }?

1 Answer

Explanation:

Related questions

Impact of this question

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