Evaluating the sum $ \sum_{n = 1}^{44} {\sin^{2}}(n^{\circ}) ~ {\cos^{2}}(n^{\circ}) $.

Question

I want to find the sum $$ {\sin^{2}}(1^{\circ}) ~ {\cos^{2}}(1^{\circ}) + {\sin^{2}}(2^{\circ}) ~ {\cos^{2}}(2^{\circ}) + {\sin^{2}}(3^{\circ}) ~ {\cos^{2}}(3^{\circ}) + \cdots + {\sin^{2}}(44^{\circ}) ~ {\cos^{2}}(44^{\circ}). $$ I thought of using the identity $ \sin(2 x) = 2 \sin(x) \cos(x) $, so $$ [2 \sin(1^{\circ}) \cos(1^{\circ})]^{2} + [2 \sin(2^{\circ}) \cos(2^{\circ})]^{2} + \cdots + [2 \sin(44^{\circ}) \cos(44^{\circ})]^{2} = 4 y. $$ By the identity above, I get $$ {\sin^{2}}(2^{\circ}) + {sin^{2}}(4^{\circ}) + \cdots + {\sin^{2}}(88^{\circ}) = 4 y, $$ but then I don’t know how to simplify this or if it could be done in other simpler ways.

Thanks.

Answer 1

Start by writing the sum in the other direction

$$\sin^288^\circ+\sin^286^\circ+...+\sin^24^\circ+\sin^22^\circ=4y$$

Now use aofkrittin's hint to rewrite this as

$$\cos^22^\circ+\cos^24^\circ+...+\cos^286^\circ+\cos^288^\circ=4y$$

Now add this to

$$\sin^22^\circ+\sin^24^\circ+...+\sin^286^\circ+\sin^288^\circ=4y$$

Summing the first terms of each equation gives $1$. Same for the second terms, third terms, fourth terms, etc. What you're left with is

$$44=8y,y=\frac{44}8=\frac{11}2$$

Answer 2

Note that $\sin^2 \theta=\frac{1-\cos 2\theta}{2}$. As there are $44$ terms in the sum, this means that

$$ 4y=\frac{44}{2}-\frac{1}{2}\left(\cos 4^\circ + \cos 8^\circ + \dots + \cos 176^\circ \right) \, . $$

Now, since $\cos(180^\circ -\theta)=-\cos \theta$, the terms in parentheses will cancel in pairs: \begin{align} \cos 4^\circ + \cos 176^\circ&= 0 \\ \cos 8^\circ + \cos 172^\circ&= 0 \\ \cdots \\ \cos 88^\circ + \cos 92^\circ&= 0 \end{align}

So the parenthesized sum vanishes, meaning that $4y=\frac{44}{2}=22$, and thus $y=\frac{11}{2}$.