4sinαsin(π/3+α)sin(π/3-α)=sin3α

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вопрос

To simplify the expression, we can use the trigonometric identity sin(A + B) = sinAcosB + cosAsinB and sin(A - B) = sinAcosB - cosAsinB.

Applying these identities, we get:

4sinαsin(π/3 + α)sin(π/3 - α)
= 4sinα(sin(π/3)cosα + cos(π/3)sinα)(sin(π/3)cosα - cos(π/3)sinα)
= 4sinα(sin(π/3)cosα + cos(π/3)sinα)(sin(π/3)cosα - sin(π/3)cosα)
= 4sinα(0 - sin^2(π/3) sinα)
= -4sinα sin^2(π/3) sinα
= -4sinα (3/2)^2 * sinα
= -4(9/4)sin^2(α)sin(α)
= -9sin^3(α)
= -sin^3(3α)
= sin3α

Therefore, the expression simplifies to sin3α.