Question

Find the value of tan(sin^-1(1/2))

Answer 1

If you know that $\sin\dfrac\pi3=\dfrac12$, then you know right away

$\tan\left(\sin^{-1}\dfrac12\right)=\tan\dfrac\pi3=\dfrac1{\sqrt}3=\dfrac{\sqrt3}3$

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Otherwise, you can derive the same result. Let $\theta=\sin^{-1}\dfrac12$, so that $\sin\theta=\dfrac12$. $\sin^{-1}$ is bounded, so we know $-\dfrac\pi2\le\theta\le\dfrac\pi2$. For these values of $\theta$, we always have $\cos\theta\ge0$.

So, recalling the Pythagorean theorem, we find

$\cos^2\theta+\sin^2\theta=1\implies\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\left(\dfrac12\right)^2}=\dfrac{\sqrt3}2$

Then

$\tan\theta=\tan\left(\sin^{-1}\dfrac12\right)=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}=\dfrac{\sqrt3}3$

as expected.

Answer 2

Answer:

c. square root 3/3

Step-by-step explanation:

just did it on edg