Let f x = sin ⁡ x + 3 sin ⁡ 3 x + 5 sin ⁡ 5 x + 3 sin ⁡ 7 x sin ⁡ 2 x + 2 sin ⁡ 4 x + 3 sin ⁡ 6 x , wherever defined. If x 1 + x 2 = π 2 , where f x is defined at x 1 and x 2 , then f 2 x 1 + f 2 x 2 is

Step-by-step solution

Numerator of f x = sin ⁡ x + sin ⁡ 3 x + 2 sin ⁡ 3 x + sin ⁡ 5 x + 3 sin ⁡ 5 x + sin ⁡ 7 x = 2 cos ⁡ x sin ⁡ 2 x + 2 sin ⁡ 4 x + 3 sin ⁡ 6 x Hence, f x = 2 cos ⁡ x , if f x 1 = 2 cos ⁡ x 1   &   f x 2 = 2 sin ⁡ x 1 i.e. f 2 x 1 + f 2 x 2 = 4