Question

the value of cos^2 5^0+cos^2 10^0+cos^2 15 +.....+cos^2 90^0=

Answer 1

Let cos(10)=a

cos(50)=b

The question becomes : a2−ab+b2

=a2−2ab+b2+ab

=(a−b)2+ab

Therefore (cos(10)−cos(50))2+cos(10)⋅cos(50)

using ; cosC−cosD=2sinC+D2sinD−C2

Therefore (2sin10+502sin50−102)2+cos(10)⋅cos(50)

=(2sin(30)sin20)2+cos(10)⋅cos(50)

=(sin20)2+cos(10)⋅cos(50)

=(sin20)2+22⋅(cos(10)⋅cos(50))

using ;

2cosacosb=cos(a+b)+cos(a−b)

=(sin20)2+cos(10+50)+cos(50−10)2

=(sin20)2+cos(60)+cos(40)2

=(sin20)2+12+cos(40)2

=(sin20)2+1+2cos(40)4

=(sin20)2+1+cos(40)2×2+cos(40)4

using ; cos2(x)=1+cos(2x)2

=(sin20)2+1+cos(40)2×2cos2(20)/2+cos(40)4

therefore