Question

Sin 3x = 3 sin x - 4 sin^3 x where, x = pie by 6

Answer 1

Alternatively, use Euler's Formula, expand cos(3x)+isin(3x) and equate the imaginary parts.

Explanation:

If you have encountered Euler's Formula then it works out like this:

Euler's Formula gives us: cosx+isinx=eix

So:

cos(3x)+isin(3x)=ei3x=(eix)3=(cos(x)+isin(x))3

=cos3(x)+3cos2(x)isin(x)+3cos(x)i2sin2(x)+i3sin3(x)

=(cos3(x)−3cos(x)sin2(x))+i(3cos2(x)sin(x)−sin3(x))

Equating the imaginary parts of both ends, we get:

sin(3x)=3cos2(x)sin(x)−sin3(x)

=sin(x)(3cos2(x)−sin2(x))

=sin(x)(3(1−sin2(x))−sin2(x))[[using: sin2(x)+cos2(x)=1]]

=sin(x)(3−4sin2(x))

=3sin(x)−4sin3(x)

Equating the real parts of both ends, we also get:

cos(3x)=cos3(x)−3cos(x)sin2(x)

=cos(x)(cos2(x)−3sin2(x))

=cos(x)(cos2(x)−3(1−cos2(x)))

=cos(x)(4cos2(x)−3)

=4cos3(x)−3cos(x)

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