Question

If 2 cos theta=x+1/x prove that 2 cos 2theta=x^2+1/x^2

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:2cos\theta = x + \dfrac{1}{x}$

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:{ \bf{ \: 2cos2\theta = {x}^{2} + \dfrac{1}{ {x}^{2} } }}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:2cos\theta = x + \dfrac{1}{x}$

Now,

Consider,

$\rm :\longmapsto\:2cos2\theta$

We know,

$\boxed{ \bf{ \: cos2x = {2cos}^{2}x - 1}}$

So, using this identity, we get

$\rm \: = \: 2 \times ({2cos}^{2}\theta - 1)$

$\rm \: = \: {4cos}^{2}\theta - 2$

$\rm \: = \: {(2cos\theta )}^{2} - 2$

We have given that,

$\rm :\longmapsto\:\boxed{ \bf{ \: 2cos\theta = x + \dfrac{1}{x} }}$

So, on substituting this value, we get

$\rm \: = \: {\bigg(x + \dfrac{1}{x} \bigg) }^{2} - 2$

We know,

$\boxed{ \bf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

So, using this identity, we get

$\rm \: = \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times \cancel{ x} \times \dfrac{1}{ \cancel{x}} - 2$

$\rm \: = \: {x}^{2} + \dfrac{1}{ {x}^{2} } + \cancel 2 - \cancel2$

$\rm \: = \: {x}^{2} + \dfrac{1}{ {x}^{2} }$

Hence,

$\bf\implies \:\boxed{ \bf{ \: 2cos2\theta = {x}^{2} + \dfrac{1}{ {x}^{2} } }}$

Additional Information :-

$\boxed{ \bf{ \: sin2x = 2sinxcosx}}$

$\boxed{ \bf{ \: sin2x = \frac{2tanx}{1 + {tan}^{2} x} }}$

$\boxed{ \bf{ \: tan2x = \frac{2tanx}{1 - {tan}^{2} x} }}$

$\boxed{ \bf{ \: cos2x = {cos}^{2}x - {sin}^{2}x}}$

$\boxed{ \bf{ \: cos2x = 1 - 2{sin}^{2}x}}$

$\boxed{ \bf{ \: cos2x = \frac{1 - {tan}^{2} x}{1 + {tan}^{2}x } }}$

$\boxed{ \bf{ \: sin3x = 3sinx - {4sin}^{3}x}}$

$\boxed{ \bf{ \: cos3x = {4cos}^{3}x - 3cosx}}$

Answer 2

Step-by-step explanation:

${x}^{2} + \frac{1}{2x}$