Squared signed sum

Using the identity $\sin(x) = \cos(\frac{\pi} {2} - x)$ , we rewrite the sum as:

$\displaystyle\sum_{n=1}^{4}(\sin^2(10n^{\circ}) + \cos^2(10n^{\circ})$ .

Using the pythagorean trigonometric identity , we get that:

$\displaystyle\sum_{n=1}^{4}(\sin^2(10n^{\circ}) + \cos^2(10n^{\circ}) = \sum_{n=1}^{4}1 = 4$ .

Therefore the value of the sum is $4$ .

A = sin^2(10)+sin^2(80) = 1 B = sin^2(20)+sin^2(70) = 1 C = sin^2(30)+sin^2(60) = 1 D = sin^2(40)+sin^2(50) = 1 sum(A+B+C+D) = 4

remember!

sin^2a+cos^2a=1 therefore sin^2a+sin^2(90-a)=1 (i call it sine couple)

there are 4 sine couples there .

so 4*1=4