Question

`sum_(k=0)^90 cos^2(pi/(180)k) =` ?

Answer 1

`cos^2(1)+...+cos^2(90^@)=45.5`

Explanation:

Remember that `cos(theta)=sin(90^@-theta)`
and that `cos(90^@)=0`

So
`cos^2(1)+....+cos^2(89)+cos^2(90)`
`color(white)("XXX")=cos^2(1)+...+cos^2(89)`
`color(white)("XXX")=sin^2(89)+...+sin^2(1)`
`color(white)("XXX")=sin^2(1)+...+sin^2(89)`

If `s=cos^2(1)+...+cos^2(89)`
then
`color(white)("XXX")2s=cos^2(1)+...+cos^2(89)`
`color(white)("XXXXX")underline(+sin^2(1)+...+sin^2(89))`
`color(white)("XXXXX")=underbrace(1color(white)("XXX")+...+color(white)("X")1)_("89 times")`

`rarr 2s =89`

`s=44.5`

Answer 2

`44.5`

Explanation:

`cos^2alpha = 1/2(cos(2alpha)+1)` so

`sum_(k=1)^n cos^2alpha_k = n/2+1/2sum_(k=1)^ncos(2alpha_k)`

now if `alpha_k = (pi/(2n))k`

`sum_(k=1)^ncos(2alpha_k) = cos(pi) = -1` (Cancellation by symmetry)

so

`sum_(k=0)^n cos^2(pi/(2n)k) = n/2-1/2 =( n-1)/2`

If `n = 90` then

`sum_(k=0)^90 cos^2(pi/(180)k) = 44.5`