How to prove this trigonometric identity?

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How to prove this trigonometric identity?

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Answer 1 · 2016-12-03T10:22:12

See below

Explanation:

$cos 2 x = {cos}^{2} x - {sin}^{2}$ $cos2x=cos^2x-sin^2$
$sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$
Are you happy about these identities?
If so then proving identities, as above, is a matter of doing something and seeing where you get to!!!
LHS can be written

$\frac{cos x - ({cos}^{2} x - {sin}^{2} x) + 2}{3 sin x - 2 sin x cos x}$ $(cosx-(cos^2x-sin^2x)+2)/(3sinx-2sinxcosx)$
= $\frac{cos x - {cos}^{2} x + {sin}^{2} x + 2}{sin x (3 - 2 cos x)}$ $(cosx-cos^2x+sin^2x+2)/(sinx(3-2cosx))$
RHS $\frac{1 + cos x}{sin x}$ $(1+cosx)/sinx$
Multiply LHS and RHS by $sin x (3 - 2 cos x)$ $sinx(3-2cosx)$
LHS= $cos x - {cos}^{2} x + {sin}^{2} x + 2$ $cosx-cos^2x+sin^2x+2$

RHS= $(1 + cos x) (3 - 2 cos x)$ $(1+cosx)(3-2cosx)$
= $3 - 2 cos x + 3 cos x - 2 {cos}^{2} x$ $3-2cosx+3cosx-2cos^2x$
= $3 + cos x - 2 {cos}^{2} x$ $3+cosx-2cos^2x$
Now looking at the LHS and remembering that ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$ or ${sin}^{2} x = 1 - {cos}^{2} x$ $sin^2x=1-cos^2x$
LHS= $cos x - {cos}^{2} x + (1 - {cos}^{2} x) + 2$ $cosx-cos^2x+(1-cos^2x)+2$
= $cos x - 2 {cos}^{2} x + 3$ $cosx-2cos^2x+3$

We are there!!!
Difficult for me to write this really clearly here, paper would be easier!!!
BUT THE IMPORTANT THING WITH IDENTITIES IS DO SOMETHING AND YOU WILL GET THERE.

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How to prove this trigonometric identity?

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