1. ## Hard Trigonometry

Evaluate the sum $\displaystyle \sin^2 1^{\circ}+\sin^2 2^{\circ}+\sin^2 3^{\circ}+\ldots+\sin^2 90^{\circ}$.

Note: The numbers are in degrees, example: $\displaystyle \sin 90^{\circ}=1$

2. Hello,

Let N be this sum.

If you replace the sinus with cosine, you'll get something interesting :

$\displaystyle \sin^2(x)=1-\cos^2(x)$

So replace for each term.

You'll have :

$\displaystyle N=(1-\cos^2 1°) + (1-\cos^2 2°) + \dots + (1-\cos^2 90°)$$\displaystyle = 90-(\cos^2 1° + \cos^2 2° + \dots + \cos^2 90°)$

But we know that $\displaystyle \cos(x)=\sin(90-x) \Longrightarrow \cos^2 (x)=\sin^2(90-x)$

So if we replace again :

$\displaystyle N=90-(\sin^2 89° + \sin^2 88° + \dots + \sin^2 0°)$ (1)

The aim will be to retrieve N in the brackets. So tihere is $\displaystyle sin^2 90°$ missing and an extra $\displaystyle \sin^2 0°$.

sin(0)=0, so we can remove $\displaystyle \sin^2 0°$. As sin(90°)=1, $\displaystyle sin^2 90°=1$

By using (1) :

$\displaystyle N=90-((\sin^2 90° + \sin^2 89° + \sin^2 88° + \dots + \sin^2 1°) - \sin^2 90°)$

$\displaystyle N=90-(N-1)$

And then you have N