Question

Brainly stars and super moderator. A question for you all:.

sin x-sin 3x / cos x+cos 3x =tan 2x
Class 11
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Answer 1

$\large \bf \clubs \:  Appropriate \: Question :-$

Prove That,

$\dfrac{ \sin\text x + \sin3\text x}{ \cos\text x + \cos3\text x} = \tan2 \text x$

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$\large \bf \clubs \:  Basic \: Formulas :-$

$\maltese \: \: \text {sin3x = 3sinx} - \text{4 {sin}}^{3} \text x \\ \\ \maltese \: \: \text{cos3x = 4{cos}}^{3} \text x - \text{3cosx} \\ \\ \maltese \: \: \text{1 - {sin}}^{2} \text{x = {cos}}^{2} \text x \\ \\ \maltese \: \: \text{2 {cos}}^{2} \text x - 1 = \text{cos2x} \\ \\ \maltese \: \: \frac{ \sin\text x}{ \cos\text x} = \tan\text x\\\\ \maltese \: \: 2\sin \text x \cos\text {x = sin2x}$

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$\large \bf \clubs \:  Proof :-$

$\large \pmb{LHS : }$

$\dfrac{ \sin\text x + \sin3\text x}{ \cos\text x + \cos3\text x}$

$= \dfrac{ \sin\text x + 3\sin\text x - 4 \sin {}^{3} \text x}{ \cos\text x + 4 { \cos}^{3}\text x - 3 \cos\text x}$

$= \dfrac{4 \sin\text x - 4 \sin {}^{3} \text x}{ 4\cos {}^{3} \text x - 2\cos\text x}$

$= \dfrac{ 4\sin\text x \: (1 - \sin {}^{2} \text x)}{ 2\cos\text x \: ( 2\cos {}^{2} \text x - 1)}$

$= \dfrac{ 2\sin\text x }{ \cancel{\cos\text x}} \times \dfrac{ \sin\text x \: { \cos}^{ \cancel2}\text x }{ \cos2\text x}$

$= \dfrac{2 \sin\text x \cos \text x}{ \cos2\text x }$

$= \dfrac{ \sin2\text x}{ \cos2\text x}$

$= \bf tan2x$

$\large =\pmb{LHS}$

$\Large\underline{\pink{\bigstar\:\:\underline{\frak{\pmb{\text Hence\:\:Proved }}}}}$

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Answer 2

☀︎︎ANSWER☀︎︎

TO PROVE:-

$\frac{sin \: x + sin \: 3x}{cos \: x + cos3x} = tan \: 2x$

It is known that,

$sin \: a \: + sin \: b \: = 2 \: sin \: ( \frac{ a + b}{2} )cos( \frac{a - b}{2} )$

$cos \: a \: + cos \: b = 2cos( \frac{a + b}{2} )cos( \frac{a - b}{2} )$

Therefore, LHS =

$\frac{sin \: x + sin \: 3x}{cos \: x + cos \: 3x}$

$= \frac{2sin( \frac{x + 3x}{2})cos( \frac{x - 3x}{2} ) }{2 \: cos( \frac{x + 3x}{2})cos( \frac{x - 3x}{2} ) }$

$= \frac{sin \: 2x}{cos \: 2x}$

$= tan \: 2x$

= R.H.S

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♥︎THANKS!!