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Mathematics Question on Trigonometric Functions of Sum and Difference of Two Angles

Find sinx2, cosx2 and tanx2 for cosx=−12 for cosx=−13,sin \frac{x}{2},\,cos \frac{x}{2} \,and \,tan\frac{ x}{2}\, for \,cos x=-\frac{1}{2} \,for\,cosx=-\frac{1}{3}, in quadrant III.

Answer

Here, x is in quadrant III.

i.e., π<x<3π2{\pi}<x<\frac{3\pi}{2}

⇒π2<x2<3π4⇒\frac{\pi}{2}<\frac{x}{2}<\frac{3\pi}{4}

Therefore, cosx2 and ,tanx2 and tanx2 are negative ,whereas  sinx2 is positive.cos\frac{x}{2}\,and\,,tan\frac{x}{2}\,and\,tan\frac{x}{2}\,\text{are\,negative\,,whereas}\,\,sin\frac{x}{2}\,is\,positive.

it is given that cosx=−13.cos x=-\frac{1}{3}.

cos x=1-2 sin2 x2\frac{x}{2}

⇒sin2x2=1−cosx2⇒sin^2\frac{x}{2}=\frac{1-cosx}{2}

⇒sin2x2=1−(−13)2=(1+13)2=432=23⇒sin^2\frac{x}{2}=\frac{1-(-\frac{1}{3})}{2}=\frac{(1+\frac{1}{3})}{2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}

⇒sinx2=23       [sinx2 is positive]⇒sin\frac{x}{2}=\frac{\sqrt2}{\sqrt3}\,\,\,\,\,\,\,[sin\frac{x}{2}\,is \,positive]

∴ sinx2=23×33=63∴\,sin\frac{x}{2}=\frac{\sqrt2}{\sqrt3}×\frac{\sqrt3}{\sqrt3}=\frac{\sqrt6}{\sqrt3}

Now, cosx=2cos2x2−1cosx=2cos^2\frac{x}{2}-1

⇒cos2x2=1+cosx2=1+(−13)2=(3−12)2=(23)2=13⇒cos^2\frac{x}{2}=\frac{1+cosx}{2}=\frac{1+(-\frac{1}{3})}{2}=\frac{(\frac{3-1}{2})}{2}=\frac{(\frac{2}{3})}{2}=\frac{1}{3}

⇒cos⁡x2=−13 [cosx2 is negative]⇒\cos\frac{x}{2}=-\frac{1}{\sqrt3}\,[cos\frac{x}{2}\,is\,negative]

∴ cosx2=−13×33=33∴\,cos\frac{x}{2}=-\frac{1}{\sqrt3}×\frac{\sqrt3}{\sqrt3}=\frac{\sqrt3}{3}

tanx2=sinx2cosx2=(23)(−13)=−2tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=\frac{(\frac{\sqrt2}{\sqrt3})}{(\frac{-1}{\sqrt3})}=-{\sqrt2}

Thus, the respective values of sinx2,cosx2 and tanx2 are 63,−33, and−2sin\frac{x}{2},cos\frac{x}{2}\,and\,tan\frac{x}{2}\,are\,\frac{\sqrt6}{3},\frac{-\sqrt3}{3},\,and-{\sqrt2}