Question #7cfc8

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Question #7cfc8

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Answer 1 · 2016-10-05T06:54:26

Proof below

Explanation:

First we will find the expansion of $sin(3x)$ separately (this will use the expansion of trig functions formulae):
$sin(3x)=sin(2x+x)$
$=sin2xcosx+cos2xsinx$
$=2sinxcosx*cosx+(cos^2x-sin^2x)sinx$
$=2sinxcos^2x+sinxcos^2x-sin^3x$
$=3sinxcos^2x-sin^3x$
$=3sinx(1-sin^2x)-sin^3x$
$=3sinx-3sin^3x-sin^3x$
$=3sinx-4sin^3x$

Now to solve the original question:
$(sin3x)/(sinx)=(3sinx-4sin^3x)/sinx$
$=3-4sin^2x$
$=3-4(1-cos^2x)$
$=3-4+4cos^2x$
$=4cos^2x-1$
$=4cos^2x-2+1$
$=2(2cos^2x-1)+1$
$=2(cos2x)+1$

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Question #7cfc8

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