Question

If $\cos x = \dfrac{{ - 1}}{3},$ x in quadrant III. Find the value of $\sin \dfrac{x}{2},\cos \dfrac{x}{2},\tan \dfrac{x}{2}$

Answer 1

Here we will first find the range of the angles for the given angle and the range of the angles for the required angle and the quadrant in which it lies. By using the trigonometric formulas and All STC rule will find out the required values.

Complete step-by-step answer:
As, given that – “x” lies in the third quadrant.
$\therefore \pi < x < \dfrac{{3\pi }}{2}$ for “x” lies in the third quadrant.
$\Rightarrow \dfrac{\pi }{2} < \dfrac{x}{2} < \dfrac{{3\pi }}{4}$for $\dfrac{x}{2}$ lies in the second quadrant.
In the second quadrant sine is positive and cosine is negative.
Now, use the identity –
$1 - \cos x = 2{\sin ^2}\dfrac{x}{2} \\\ \Rightarrow \sin \dfrac{x}{2} = \pm \sqrt {\dfrac{{1 - \cos x}}{2}} \\\$
Place the values in the above equation-
$\Rightarrow \sin \dfrac{x}{2} = \pm \sqrt {\dfrac{{1 - \left( {\dfrac{{ - 1}}{3}} \right)}}{2}}$
Simplify the above equation –
$\Rightarrow \sin \dfrac{x}{2} = \pm \sqrt {\dfrac{4}{6}} \\\ \Rightarrow \sin \dfrac{x}{2} = \pm \dfrac{{\sqrt 2 }}{{\sqrt 3 }} \\\$
Since the above angle is in the second quadrant.
$\therefore \sin \dfrac{x}{2} = \sqrt {\dfrac{2}{3}} {\text{ }}....{\text{ (A)}}$
Now, use the identity –
$1 + \cos x = 2co{\operatorname{s} ^2}\dfrac{x}{2} \\\ \Rightarrow \cos \dfrac{x}{2} = \pm \sqrt {\dfrac{{1 + \cos x}}{2}} \\\$
Place the values in the above equation-
$\Rightarrow \cos \dfrac{x}{2} = \pm \sqrt {\dfrac{{1 + \left( {\dfrac{{ - 1}}{3}} \right)}}{2}}$
Simplify the above equation –
$\Rightarrow \cos \dfrac{x}{2} = \pm \sqrt {\dfrac{1}{3}} \\\ \Rightarrow \cos \dfrac{x}{2} = \pm \dfrac{1}{{\sqrt 3 }} \\\$
Since the above angle is in the second quadrant, cosine is negative
$\therefore \cos \dfrac{x}{2} = - \dfrac{1}{{\sqrt 3 }}{\text{ }}....{\text{ (B)}}$
Now, use identity –
$\tan \dfrac{x}{2} = \dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}$
Place values in the above equation-
$\tan \dfrac{x}{2} = \dfrac{{\dfrac{{\sqrt 2 }}{{\sqrt 3 }}}}{{ - \dfrac{1}{{\sqrt 3 }}}}$
Like and same terms from the denominator and the numerator cancel each other. So remove them and simplify the fraction.
$\Rightarrow \tan \dfrac{x}{2} = - \sqrt 2 {\text{ }}....{\text{ (C)}}$

Note: Remember the trigonometric formulas and the correlation between the trigonometric functions to find the unknowns. Also, remember the All STC rule, it is also known as the ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ($0^\circ \;{\text{to 90}}^\circ$ ) are positive, sine and cosec are positive in the second quadrant ($90^\circ {\text{ to 180}}^\circ$ ), tan and cot are positive in the third quadrant ($180^\circ \;{\text{to 270}}^\circ$ ) and sin and cosec are positive in the fourth quadrant ($270^\circ {\text{ to 360}}^\circ$ ).