Question #9c7b3

2 Answers
Cesareo R.
Nov 25, 2016

#sin^3x=(3sinx-sin(3x))/4#

Explanation:

If you know the de Moivre's identity it is quite direct.

#e^(i x) = cosx+isinx#

so

#e^(3ix)= (cosx + isinx)^3=cos(3x)+isin(3x)#

or

#cos^3x+3icos^2xsinx-3cosxsin^2x-isin^3x=cos(3x)+isin(3x)#

grouping the real and the imaginary parts,

#cos^3x-3cosxsin^2x=cos(3x)# and
#3cos^2xsinx-sin^3x=sin(3x)#

now, using the identity

#cos^2x+sin^2x=1# we will get at

#sin(3x)=3sinx(1-sin^2x)-sin^3x# or

#sin^3x=(3sinx-sin(3x))/4#

Nghi N
Nov 25, 2016

Expand #sin^3 x#

Explanation:

There is another simpler method.
Use the trig identity:
#sin 3x = 3sin x - 4sin^3 x#
By transposing terms, we get:
#4sin^3 x = 3sin x - sin 3x#
#sin^3 x = (3sin x - sin 3x)/4#

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