# Math Help - Exact Values

1. ## Exact Values

Hey i need help finding exact values for 2 problems

1) sin[2sin^-1(1/2)]

2) sec[tan^-1(1/2)]

2. Originally Posted by DjamaalRobinson
Hey i need help finding exact values for 2 problems

1) sin[2sin^-1(1/2)]
I do this one, $\sin^{-1} \tfrac{1}{2} = \tfrac{\pi}{6}$.
Thus, $\sin [2\cdot \tfrac{\pi}{6}] = \sin \tfrac{\pi}{3} = \tfrac{\sqrt{3}}{2}$.

3. Originally Posted by DjamaalRobinson
Hey i need help finding exact values for 2 problems

1) sin[2sin^-1(1/2)]

2) sec[tan^-1(1/2)]
1) $\sin\left[2\sin^{-1}\left(\tfrac{1}{2}\right)\right]$

Depending on the restriction on $\vartheta$, we see that $\vartheta=\sin^{-1}\left(\tfrac{1}{2}\right)=\frac{\pi}{6}~or~\frac {5\pi}{6}$

Thus, we can either have $\sin\left(2\cdot\frac{\pi}{6}\right)=\sin\left(\fr ac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$, or $\sin\left(2\cdot\frac{5\pi}{6}\right)=\sin\left(\f rac{5\pi}{3}\right)=-\frac{\sqrt{3}}{2}$

2) $\sec\left[\tan^{-1}\left(\tfrac{1}{2}\right)\right]$

Construct a triangle to represent the angle $\vartheta=\tan^{-1}\left(\tfrac{1}{2}\right)$

This implies that we have an opposite of length 1, and an adjacent of length 2. Thus, the length of the hypoteuse can be found by applyin the pythagorean forumla:

$h^2=1+4\implies h=\sqrt{5}$

Thus, $\sec\vartheta=\color{red}\boxed{\frac{\sqrt{5}}{2} }$

I hope this makes sense!

--Chris

4. Originally Posted by Chris L T521
Depending on the restriction on $\vartheta$
The arcsine function is defined with range $[-\pi/2,\pi/2]$