# Find tan(x/2) from the given information. cos(x)= -24/25, 180 < x < 270

## Question:

Find {eq}\tan \frac{x}{2} {/eq} from the given information.

{eq}\cos x= -\frac{24}{25}, 180^\circ < x < 270^\circ {/eq}

## Tangent of an angle:

In mathematics, the tangent of an angle can be defined as the ratio of the sine of the angle to the cosine of the angle. The expression for the tangent of an angle is,

{eq}\tan x = \dfrac{{\sin x}}{{\cos x}} {/eq}

Here, sine of an angle is {eq}\sin x {/eq} and the cosine of an angle is {eq}\cos x {/eq}.

Given Data

• The value for {eq}\cos x {/eq} is {eq}\cos x = - \dfrac{{24}}{{25}} {/eq}.

The interval for {eq}x {/eq} is {eq}180^\circ < x < 270^\circ {/eq} thus, the interval for {eq}{\frac{x}{2}} {/eq} is {eq}90^\circ < x < 135^\circ {/eq} this interval lies in the second quadrant.

For the given the interval the value for {eq}\sin \left( {\dfrac{x}{2}} \right) {/eq} is positive and the value for the {eq}\cos \left( {\dfrac{x}{2}} \right) {/eq} is negative.

The expression for the {eq}\sin \left( {\dfrac{x}{2}} \right) {/eq} by the Half-Angle Rule is,

{eq}\sin \dfrac{x}{2} = \sqrt {\dfrac{{1 - \cos x}}{2}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ \begin{align*} \cos 2x &= 1 - 2{\sin ^2}x\\ {\sin ^2}x &= \dfrac{{1 - \cos 2x}}{2}\\ \sin x &= \sqrt {\dfrac{{1 - \cos 2x}}{2}} \end{align*} \right] {/eq}

Substitute the given values.

{eq}\begin{align*} \sin \dfrac{x}{2} &= \sqrt {\dfrac{{1 - \left( { - \dfrac{{24}}{{25}}} \right)}}{2}} \\ &= \sqrt {\dfrac{{\dfrac{49}{{25}}}}{2}} \\ &= \sqrt {\dfrac{49}{{50}}} \\ &=\pm \dfrac{7}{{5\sqrt 2 }} &\text{which is + for our given interval}\\ &= \dfrac{7\sqrt{2}}{{10}} \end{align*} {/eq}

The expression for the {eq}\cos \left( {\dfrac{x}{2}} \right) {/eq} by the Half-Angle Rule is,

{eq}\cos \dfrac{x}{2} = \sqrt {\dfrac{{1 + \cos x}}{2}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ \begin{align*} \cos 2x &= 2{\cos ^2}x - 1\\ {\cos ^2}x &= \dfrac{{1 + \cos 2x}}{2}\\ \cos x &= \sqrt {\dfrac{{1 + \cos 2x}}{2}} \end{align*} \right] {/eq}

Substitute the given values.

{eq}\begin{align*} \cos \dfrac{x}{2} &= \sqrt {\dfrac{{1 + \left( { - \dfrac{{24}}{{25}}} \right)}}{2}} \\ &= \sqrt {\dfrac{{\dfrac{1}{{25}}}}{2}} \\ &= \sqrt {\dfrac{1}{{50}}} \\ &= \pm \dfrac{1}{{5\sqrt 2 }} &\text{which is - for our given interval}\\ &= -\dfrac{\sqrt{2}}{{10}} \end{align*} {/eq}

The expression for the {eq}\tan \left( {\dfrac{x}{2}} \right) {/eq} is,

{eq}\tan \left( {\dfrac{x}{2}} \right) = \dfrac{{\sin \left( {\dfrac{x}{2}} \right)}}{{\cos \left( {\dfrac{x}{2}} \right)}} {/eq}

Substitute the given values.

{eq}\begin{align*} \tan \left( {\dfrac{x}{2}} \right) &= \dfrac{{\dfrac{7\sqrt{2}}{{10}}}}{{-\dfrac{\sqrt{2}}{{10}}}}\\ &= -\dfrac{7}{1}\\ &= - 7 \end{align*} {/eq}

Thus, the value for {eq}\tan \left( {\dfrac{x}{2}} \right) {/eq} is -7.