The value of cos(x/2) is required.
The required value is [tex]\cos(x/2)=-\sqrt{\dfrac{5-2\sqrt{5}}{10}}[/tex]
Trigonometry
The given value is
[tex]\tan x=\dfrac{1}{2}[/tex] where x is in quadrant III.
From trigonometric identities we have
[tex]\sec^2x=1+\tan^2x\\\Rightarrow \sec^2x=1+(\dfrac{1}{2})^2\\\Rightarrow \sec^2x=\dfrac{5}{4}\\\Rightarrow \cos^{2}x=\dfrac{4}{5}\\\Rightarrow \cos x=\dfrac{2}{\sqrt{5}}[/tex]
If x lies in quadrant III then x/2 lies in quadrant II.
Cos is negative in quadrant III so,
[tex]\cos x=-\dfrac{2}{\sqrt{5}}=-\dfrac{2\sqrt{5}}{5}[/tex]
[tex]\cos \dfrac{x}{2}=\pm \sqrt{\dfrac{1+\cos x}{2}}\\ =\pm\sqrt{\dfrac{1-\dfrac{2\sqrt{5}}{5}}{2}}\\ =\pm\sqrt{\dfrac{5-2\sqrt{5}}{10}}[/tex]
Since, x/2 lies in the second quadrant and cos (x/2) is negative.
The required value is [tex]-\sqrt{\dfrac{5-2\sqrt{5}}{10}}[/tex]
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