Question

If $\sin \theta =\frac{-1}{3}$ and $\theta$ does not lie in third quadrant, find the values of $\cos \theta$.

Answer 1

$\sin \theta =-\frac{1}{3}=>\sin {}^2\theta =\frac{1}{9}\sin {}^2\theta +\cos {}^2\theta =1\frac{1}{9}+\cos {}^2\theta =1\cos {}^2\theta =1-\frac{1}{9}=\frac{8}{9}\cos \theta =\pm \sqrt{\frac{8}{9}}=\pm \frac{2\sqrt{2}}{3}=>\cos \theta =\frac{2\sqrt{2}}{3}or\cos \theta =-\frac{2\sqrt{2}}{3}$

Now $\sin \theta$ is negative, $\theta$ must lie in $3$rd or $4$th quadrant.

But it is given that $\theta$ does not lie in $3$rd quadrant..

So $\theta$ lies in $4$th quadrant

In $4$th quadrant $\cos \theta$ is positive.

$=>\cos \theta =-\frac{2\sqrt{2}}{3}=>\cos \theta =\frac{2\sqrt{2}}{3}Ans=\frac{2\sqrt{2}}{3}$