Question

if cosx = -3/5, x lies in II quadrant . find the value sin x/2, cosx/2 and tan x /2

Answer 1

Question :-

• if cosx = -3/5, x lies in II quadrant . find the value sin x/2, cosx/2 and tan x /2

Answer

Given :-

• cosx = -3/5, x lies in II quadrant .

To find :-

• The value sin x/2, cosx/2 and tan x /2

Identity used :-

$\bf \:cosx = 1 - 2 {sin}^{2} \dfrac{x}{2}$

$\bf \:cosx = 2 {cos}^{2} \dfrac{x}{2} - 1$

Solution :-

As x lies in second quadrant.

$\bf\implies \:\dfrac{\pi}{2} < x < \pi$

$\bf\implies \:\dfrac{\pi}{4} < \dfrac{x}{2} < \dfrac{\pi}{2}$

$\bf\implies \:\dfrac{x}{2} \: lies \: in \: first \: quadrant.$

$\bf\implies \:sin\dfrac{x}{2} , \: cos\dfrac{x}{2}, \: tan\dfrac{x}{2} > 0$

(1) To find the value of sin(x/2), we know

$\bf \:cosx = 1 - 2 {sin}^{2} \dfrac{x}{2}$

Put the value of cosx, we get

$\bf \: - \dfrac{3}{5} = 1 - 2 {sin}^{2} \dfrac{x}{2}$

$\bf\implies \:2 {sin}^{2} \dfrac{x}{2} = 1 + \dfrac{3}{5}$

$\bf\implies \:2 {sin}^{2} \dfrac{x}{2} = \dfrac{8 }{5}$

$\bf\implies \: {sin}^{2} \dfrac{x}{2} = \dfrac{4}{5}$

$\bf\implies \:sin\dfrac{x}{2} = \dfrac{2}{ \sqrt{5} }$

(2) To find the value of cos(x/2), we know

$\bf \:cosx = 2 {cos}^{2} \dfrac{x}{2} - 1$

Put the value of cosx, we get

$\bf \: - \dfrac{3}{5} = 2 {cos}^{2} \dfrac{x}{2} - 1$

$\bf\implies \: {2cos}^{2} \dfrac{x}{2} = 1 - \dfrac{3}{5}$

$\bf\implies \: {2cos}^{2} \dfrac{x}{2} = \dfrac{2}{5}$

$\bf\implies \: {cos}^{2} \dfrac{x}{2} = \dfrac{1}{5}$

$\bf\implies \:cos\dfrac{x}{2} = \dfrac{1}{ \sqrt{5} }$

(3) To find the value of tan(x/2)

$\bf \:tan \dfrac{x}{2} = \dfrac{sin\dfrac{x}{2}}{cos\dfrac{x}{2}}$

$\bf\implies \:tan\dfrac{x}{2} = \dfrac{\dfrac{2}{ \sqrt{5} }}{ \dfrac{1}{ \sqrt{5} }}$

$\bf\implies \:tan\dfrac{x}{2} = 2$

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