cos(π-α) + cos(3π/2+α) = -sinα - sinα = -2sinα
Теперь выражаем sinα через cos:
cos(π-α) + cos(3π/2+α) = -2sinα = -2(sqrt(1-cos^2(α)))
cos(π-α) + cos(3π/2+α) = -2(sqrt(1-cos^2(α)))
cos(π-α) + cos(3π/2+α) = -2(sqrt((1+cosα)(1-cosα)))
cos(π-α) + cos(3π/2+α) = -2(sqrt(1-cos^2(α))) = -2(sqrt(1-sin^2(α)))
cos(π-α) + cos(3π/2+α) = -2(sqrt(1-cos^2(α))) = -2cosα
Ответ: -2cosα
cos(π-α) + cos(3π/2+α) = -sinα - sinα = -2sinα
Теперь выражаем sinα через cos:
cos(π-α) + cos(3π/2+α) = -2sinα = -2(sqrt(1-cos^2(α)))
cos(π-α) + cos(3π/2+α) = -2(sqrt(1-cos^2(α)))
cos(π-α) + cos(3π/2+α) = -2(sqrt((1+cosα)(1-cosα)))
cos(π-α) + cos(3π/2+α) = -2(sqrt(1-cos^2(α))) = -2(sqrt(1-sin^2(α)))
cos(π-α) + cos(3π/2+α) = -2(sqrt(1-cos^2(α))) = -2cosα
Ответ: -2cosα