c o s 2 θ cos^2 \theta series

Geometry Level 2

c o s 2 1 5 ∘ + c o s 2 2 0 ∘ + c o s 2 2 5 ∘ + . . . . . . . . . . . . + c o s 2 7 5 ∘ = ? ? \large cos^2 15^{\circ}+cos^2 20^{\circ}+cos^2 25^{\circ}+............+cos^2 75^{\circ}=??

Find the value of that series.

6 0 None of them 13 2 \frac{13}{2} ∞ \infty

Md Mehedi Hasan by Md Mehedi Hasan , 22, Bangladesh

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1 solution

Md Mehedi Hasan
Nov 25, 2017

c o s 2 1 5 ∘ + c o s 2 2 0 ∘ + c o s 2 2 5 ∘ + . . . . . . . . . . . . + c o s 2 7 5 ∘ = c o s 2 1 5 ∘ + c o s 2 2 0 ∘ + c o s 2 2 5 ∘ + . . . . . . . . . . . . + c o s 2 7 0 ∘ + c o s 2 7 5 ∘ = c o s 2 1 5 ∘ + c o s 2 2 0 ∘ + c o s 2 2 5 ∘ + . . . . . . . . . . . . + c o s 2 ( 90 − 20 ) ∘ + c o s 2 ( 90 − 15 ) ∘ = c o s 2 1 5 ∘ + c o s 2 2 0 ∘ + c o s 2 2 5 ∘ + . . . . . . . . . . . . + s i n 2 2 0 ∘ + s i n 2 1 5 ∘ cos^2 15^{\circ}+cos^2 20^{\circ}+cos^2 25^{\circ}+............+cos^2 75^{\circ}\\ =cos^2 15^{\circ}+cos^2 20^{\circ}+cos^2 25^{\circ}+............+cos^2 70^{\circ}+cos^2 75^{\circ}\\ =cos^2 15^{\circ}+cos^2 20^{\circ}+cos^2 25^{\circ}+............+cos^2(90-20)^{\circ} +cos^2 (90-15)^{\circ}\\ =cos^2 15^{\circ}+cos^2 20^{\circ}+cos^2 25^{\circ}+............+sin^2 20^{\circ}+sin^2 15^{\circ}

= ( c o s 2 1 5 ∘ + s i n 2 1 5 ∘ ) + ( c o s 2 2 0 ∘ + s i n 2 2 0 ∘ ) + . . . . . . . . . . . + c o s 2 4 5 ∘ Here are 6 pair. We can easily find it with a+(n-1)d formula = 1 + 1 + 1 + 1 + 1 + 1 + 1 2 = 13 2 =(cos^2 15^{\circ}+sin^2 15^{\circ})+(cos^2 20^{\circ}+sin^2 20^{\circ})+...........+cos^2 45^{\circ}\quad\boxed{\color{#3D99F6}{\text{Here are 6 pair. We can easily find it with a+(n-1)d formula}}}\\ =1+1+1+1+1+1+\frac{1}{2}\\ =\boxed{\frac{13}{2}}

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