Question

If tanx = -3/4 and x lies in fourth quadrant, find sinx/2, cosx/2, tanx/2​

Answer 1

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

Given that,

$\rm \: tanx = - \frac{3}{4} \: \: and \: x \: lies \: in \: {4}^{th} \: quadrant$

Now, We know that,

$\rm \: {sec}^{2}x - {tan}^{2}x = 1$

On substituting the value of tanx, we get

$\rm \: {sec}^{2}x - {\bigg( - \dfrac{3}{4} \bigg) }^{2}= 1$

$\rm \: {sec}^{2}x - \frac{9}{16} = 1$

$\rm \: {sec}^{2}x = 1 + \frac{9}{16}$

$\rm \: {sec}^{2}x = \frac{25}{16}$

$\rm \: secx \: = \: \pm \: \frac{5}{4}$

As it is given that, x lies in 4th quadrant, so secx > 0

$\rm\implies \:secx = \frac{5}{4}$

$\rm\implies \:cosx = \frac{4}{5}$

As it is given that x lies in 4th quadrant.

$\rm \: \frac{3\pi}{2} < x < 2\pi$

$\rm\implies \:\rm \: \frac{3\pi}{4} < \frac{x}{2} < \pi$

$\rm\implies \: \frac{x}{2} \: lies \: in \: {2}^{nd} \: quadrant$

$\rm\implies \:sin \frac{x}{2} > 0, \: \: \: and \: \: \: \: cos \frac{x}{2} < 0 \:$

Now, we know that

$\rm \: cosx = {2cos}^{2} \frac{x}{2} - 1$

On substituting the value of cosx, we get

$\rm \: \frac{4}{5} = {2cos}^{2} \frac{x}{2} - 1$

$\rm \: \frac{4}{5} + 1= {2cos}^{2} \frac{x}{2}$

$\rm \: \frac{4 + 5}{5}= {2cos}^{2} \frac{x}{2}$

$\rm \: \frac{9}{10}= {cos}^{2} \frac{x}{2}$

$\rm \: cos \frac{x}{2} \: = \: \pm \: \frac{3}{ \sqrt{10} }$

As,

$\rm \: cos \frac{x}{2} \: < 0$

$\rm\implies \:\rm \: cos \frac{x}{2} \: = \: - \: \frac{3}{ \sqrt{10} }$

Also, we know, that

$\rm \: cosx = 1 - {2sin}^{2} \frac{x}{2}$

So, on substituting the value of cosx, we get

$\rm \: \frac{4}{5} = 1 - {2sin}^{2} \frac{x}{2}$

$\rm \: \frac{4}{5} - 1 = {2sin}^{2} \frac{x}{2}$

$\rm \: \frac{4 - 5}{5} = {2sin}^{2} \frac{x}{2}$

$\rm \: - \: \frac{1}{5} = - \: {2sin}^{2} \frac{x}{2}$

$\rm \: \: \frac{1}{10} = \: {sin}^{2} \frac{x}{2}$

$\rm \: sin \frac{x}{2} \: = \: \pm \: \frac{1}{ \sqrt{10} }$

As,

$\rm \: sin \frac{x}{2} \: > 0$

So,

$\rm\implies \:\rm \: sin \frac{x}{2} \: = \: \frac{1}{ \sqrt{10} }$

Now,

$\rm \: tan \frac{x}{2}$

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

$\rm \: = \: \frac{1}{ \sqrt{10} } \div \bigg( - \frac{ 3 }{10} \bigg)$

$\rm \: = \: - \: \frac{1}{3}$

Hence,

$\rm\implies \:\rm \: cos \frac{x}{2} \: = \: - \: \frac{3}{ \sqrt{10} }$

$\rm\implies \:\rm \: sin \frac{x}{2} \: = \: \frac{1}{ \sqrt{10} }$

$\rm\implies \:\rm \: tan \frac{x}{2} \: = \: \frac{1}{3}$

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Additional Information :-

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

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

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

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

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

$\boxed{\sf{ \:tan3x = \frac{3tanx - {tan}^{3} x}{1 - 3 {tan}^{2} x} \: \: }}$