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

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

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Question #7cfc8

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