Let x_n = ^2(n^ ) . What is the mean of x_1,x_2,x_3, ,x_90 ?

TMUA practice: Trig Identities, problem 2

A TMUA-calibre trig identities problem at difficulty 4 of 5, with the full worked solution.

The question

Let

x_n = \sin^2(n^{\circ})

. What is the mean of

x_1,x_2,x_3,\dots,x_{90}

\frac{11}{45}

\frac{22}{45}

\frac{89}{180}

\frac{1}{2}

\frac{91}{180}

The correct answer is highlighted. E

Worked solution

Add up $x_1$ with $x_{89}$ , $x_2$ with $x_{88}$ , and $x_i$ with $x_{90-i}$ . Notice

x_i+x_{90-i}=\sin^2(i^{\circ})+\sin^2((90-i)^{\circ})=\sin^2(i^{\circ})+\cos^2(i^{\circ})=1

by the Pythagorean identity. Since we can pair up $1$ with $89$ and keep going until $44$ with $46$ , we get

x_1+x_2+\dots+x_{90}=44+x_{45}+x_{90}=44+\left(\frac{\sqrt{2}}{2}\right)^2+1^2=\frac{91}{2}

Hence the mean is $\boxed{\textbf{(E) }\frac{91}{180}}$