sin²5° + sin²10° + sin²15° + ...... sin²85° + sin²90° is equal to?

Solution(By Examveda Team)

Given, (sin²5° + sin²10° + sin²15° + . . . . . . + sin²85° ) + sin²90°
We know that sinθ = cos(90° - θ )
Therefore sin²85° = cos²(90° - 85°) = cos²5°
Similary sin²60° = cos²(90° - 60°) = cos²40°
And we also know that sin²θ + cos²θ = 1
There are 8 pair in given equation
sin²5° + cos²5° = 1
sin²10° + cos²10° = 1
. . . . . . . . . . . .
. . . . . . . . . . . .
sin²40° + cos²40° = 1
i.e. 8 + sin²45° + sin²90°
⇒ 8 + $$\frac{1}{2}$$ + 1 = $$9\frac{1}{2}$$

Short Trick:
 $$\left( {si{n^2}{5^ \circ }\, + \,{{\sin }^2}{{10}^ \circ }\, + \,{{\sin }^2}{{15}^ \circ }\, + \,.....\,{{\sin }^2}{{85}^ \circ }} \right)\, + $$         $$\,{\sin ^2}90$$
$$\eqalign{ & {\text{Number of terms}} \cr & = \left\{ {\left( {\frac{{85 - 5}}{5}} \right) + 1} \right\}{\text{ + 1}} \cr & {\text{ = }}\frac{{17}}{2}{\text{ + 1}} \cr & {\text{Sum of series}} = 9\frac{1}{2} \cr} $$