Question

What is the value of  sinθ, given that cosθ=−2/5 and  sinθ>0? Simplify radicals and rationalize the denominator, if necessary.   Enter your answer in the box. sinθ=

Answer 1

$\cos\theta=-\dfrac25$
$\sin\theta=\pm\sqrt{1-\cos^2\theta}=\pm\dfrac{\sqrt{21}}5$

Take the positive root, since it's given that $\sin\theta>0$.

$\sin\theta=\dfrac{\sqrt{21}}5$