Помогите пожалуйста решить. 5 cos(π/2-α), если cos α =1/8 и π<α<3π/2

5 cos(π/2-α) = 5 cos(π/2-(π/2-arccos(1/8))) = 5 cos(arccos(1/8)) = 5 (1/8) = 5/8.

5 cos(π/2-α) = 5 cos (π/2-(π/2 - arccos(1/8))) 5 cos(π/2-α) = 5 cos (arccos(1/8)) 5 cos(π/2-α) = 5 × 1/8 5 cos(π/2-α) = 5/8