Question

sin? 1° + sin? 2° + ......
+ sin290º =​

Answer 1

Answer:

We have to find the sum of the series (sin 1)^2 + (sin 2)^2 +...+ (sin 90)^2

(sin 1)^2 + (sin 2)^2 +...+ (sin 90)^2

=> (sin 1)^2 + (sin 2)^2 +...+ (sin 90)^2

=> (sin 1)^2 + (sin 2)^2 +... + (sin 44)^2 + (sin 45)^2 + (sin 46)^2 ...+ (sin 90)^2

=> (sin 1)^2 + (sin 2)^2 +... + (sin 44)^2 + (sin 45)^2 + (sin 46)^2 ...+ (sin 90)^2

use sin (90 - x) = cos x

=> (sin 1)^2 + (sin 2)^2 +... + (sin 44)^2 + (sin 45)^2 + (sin(90 - 44))^2 ...+ (sin (90 - 0))^2

=> (sin 1)^2 + (sin 2)^2 +... + (sin 44)^2 + (sin 45)^2 + (cos 44)^2 + (cos 43)^2 ...+ (cos 0)^2

=> (sin 1)^2 + (cos 1)^2 + (sin 2)^2 + (cos 2)^2 +...+ (sin 44)^2 + (cos 44)^2 + (sin 45)^2 + (cos 0)^0

=> 44*1 + (1/sqrt 2)^2 + 1

=> 44 + 1/2 + 1

=> 45.5

The required sum is 45.5

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