Question

$\sin^{2}1 + \sin^{2}3 + \sin^{2}5 + ....... + \sin^{2}85 + \sin^{2}87 + \sin^{2}89 \: = \: 22.5 \: \: \: proof \: that$

Answer 1

Use the identity
$\sin{}^{2}(x) = 1 - { \cos(x) }^{2}$
When we do that, we get
${ \sin}^{2} (3) = 1 - { \cos }^{2} (3)$
Now using the identity
$\cos( \alpha ) = \sin(90 - \alpha )$
We get
$1 - { \cos }^{2} 3 = 1 - { \sin}^{2} 87$
Now substituting that, we see that sin square 87 gets cancelled. Similarly, all the sin squares get cancelled, except for sin square 45.
Now the question reduces to
$1 + 1 + 1...... + { \sin }^{2} 45$
Now there are 22 numbers from 1 to 43 except 45, therefore
$= 22 +({ \frac{1}{ \sqrt{2} } })^{2} \\ = 22 + \frac{1}{2} \\ = 22.5$