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By writing sin 3x = sin(2x+x), show that sin 3x = 3 sin x - 4 sin^3 x.

Sagot :

By writing sin 3x = sin(2x+x), the result of the expression sin 3x = 3 sin x - 4 sin^3 x is true as shown in the proof.

Given the trigonometry expression:

sin 3x = sin(2x+x),

In trigonometry identity;

[tex]Sin(A+B) = sinAcosB + cosAsinB[/tex]

The expression above will then become;

[tex]sin(2x+x) =sin2xcosx+cos2xsinx\\sin(2x+x) =2sinxcosxcosx + (1-2sin^2x)sinx\\sin(2x+x) =2sinxcos^2x + (1-2sin^2x)sinx\\[/tex]

Recall that sin^2x + cos^x = 1, hence;

[tex]sin(2x+x) =2sinx(1-sin^2x)+ (1-2sin^2x)sinx\\Expand\\sin(2x+x) =2sinx-2sin^3x+ sinx-2sin^3x\\sin(2x+x) =2sinx + sinx-2sin^3x-2sin^3x\\sisn(2x+x) = 3sinx-4sin^3x[/tex]

This shows that [tex]sin3x =3sinx-4sin^3x[/tex] (Proved!)

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