Question

If cosx=-(⅓) x lies in third quadrant then cos(x/2) will be (a)-1/✓3 (b)1/✓3 (c)-1/6 (d)1/6

Answer 1

Given:- cosx=-(⅓) ,x lies in third quadrant

To find:- the value of cos(x/2)

Solution:-

cosx=-(⅓) ,where x lies in third quadrant.

To get the value of cos(x/2) we need to use the equation 1+ cosx = 2cos²x/2

$1 + cosx = 2 {cos}^{2} \frac{x}{2}$

$or, \frac{1 + cosx}{2} = {cos}^{2} \frac{x}{2}$

$or, cos \frac{x}{2} = ± \sqrt{ \frac{1 + cosx}{2} }$

Now we get the value of cos(x/2).

We know that, cosx = -(⅓)

$cos \frac{x}{2} = ± \sqrt{ \frac{1 + ( - \frac{1}{3} )}{2} }$

$cos \frac{x}{2} = ± \sqrt{ \frac{1}{3} }$

$\pi < x < \frac{3\pi}{2} = > \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$

and cos is negative in second quadrant.

$∴cos \frac{x}{2} = - \sqrt{ \frac{1}{3} }$

Hence ,the option (a) - 1/√3 is correct.(Ans)

Answer 2

SOLUTION

GIVEN

$\displaystyle \sf{} \cos x = - \frac{1}{3} \: \: \: \: and \: \:$

x lies in third quadrant

TO CHOOSE THE CORRECT OPTION

$\displaystyle \sf{} \cos \frac{x}{2} =$

$\displaystyle \sf{}(a) \: \: - \frac{1}{ \sqrt{3} }$

$\displaystyle \sf{}(b) \: \: \frac{1}{ \sqrt{3} }$

$\displaystyle \sf{}(c) \: \: - \frac{1}{6 }$

$\displaystyle \sf{}(d) \: \: \frac{1}{ 6 }$

FORMULA TO BE IMPLEMENTED

We are aware of the Trigonometric formula that

$\displaystyle \sf{} \cos x \: = \: 2 {\cos}^{2} \frac{x}{2} - 1$

EVALUATION

Here it is given that x lies in third quadrant

So

$\displaystyle \sf{} \: {180}^{ \circ} < x < {270}^{ \circ}$

$\implies \displaystyle \sf{} \: {90}^{ \circ} < \frac{x}{2} < {135}^{ \circ}$

$\therefore \: \: \displaystyle \sf{} \frac{x}{2} \: \: \: lies \: in \: second \: quadrant$

$\therefore \displaystyle \sf{} \: \: \cos\frac{x}{2} \: \: is \: negative$

Now

$\displaystyle \sf{} \cos x = - \frac{1}{3}$

$\implies \displaystyle \sf{}2 {\cos}^{2} \frac{x}{2} - 1 = - \frac{1}{3}$

$\implies \displaystyle \sf{}2 {\cos}^{2} \frac{x}{2} = 1 - \frac{1}{3}$

$\implies \displaystyle \sf{}2 {\cos}^{2} \frac{x}{2} = \frac{2}{3}$

$\implies \displaystyle \sf{} {\cos}^{2} \frac{x}{2} = \frac{1}{3}$

$\implies \displaystyle \sf{} {\cos} \frac{x}{2} = \pm \: \frac{1}{ \sqrt{3} }$

$\because \displaystyle \sf{} \: \: \cos\frac{x}{2} \: \: is \: negative$

$\therefore \displaystyle \sf{} \: \: \cos\frac{x}{2} = - \frac{1}{ \sqrt{3} }$

FINAL ANSWER

Hence the correct option is

$\displaystyle \sf{} \cos \frac{x}{2} =$

$\displaystyle \sf{}(a) \: \: - \frac{1}{ \sqrt{3} }$

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