I need help with a few trig problems

[kolaym]

I have a question on a few trig problems!

what is cos inverse of x= pie/3 find the sin inverse of x

 

[Unco]

G'day, kolaym.

If \(\displaystyle \L \cos^{-1}{x} = \frac{\pi}{3}\)

then \(\displaystyle \L x = \cos{\frac{\pi}{3}}\)

or \(\displaystyle \L \cos{\frac{\pi}{3}} = \frac{x}{1}\)

and we can construct the right-angled triangle:

                 /|
                  |
          1  /    |
                  |
         /        |        
         pi/3     |
     /____________|
            x
 
     as cos(pi/3) = x/1 = adjacent/hypotenuse

And if \(\displaystyle \L sin^{-1}{x} = \theta\)
where we want to know \(\displaystyle \L \theta\),

we can write as \(\displaystyle \L \sin{\theta} = x = \frac{x}{1}\)

and look at the triangle to see that the other angle, namely \(\displaystyle \L \frac{\pi}{6}\), will be \(\displaystyle \L \theta\)

                 /|
                pi|
          1  / ---|
                6 |
         /        |        
         pi/3     |
     /____________|
            x

as \(\displaystyle \L \sin{\frac{\pi}{6}} = \frac{opp}{hyp} = \frac{x}{1} = x\)

That is, \(\displaystyle \L \sin^{-1}{x} = \frac{\pi}{6}\)

 

[soroban]

Hello, kolaym!

Very nice explanation, Unco!
Here's an "eyeball" approach . . . well, almost.

\(\displaystyle \text{If }\cos^{-1}(x)\,=\,\frac{\pi}{3},\,\text{ find }\sin^{-1}(x)\)

We have: \(\displaystyle \L\,\cos^{-1}(x)\,=\,\frac{\pi}{3}\;\;\Rightarrow\;\;x \,=\,\cos(\frac{\pi}{3}) \,=\,\frac{1}{2}\)


They asked for: \(\displaystyle \L\,\sin^{-1}\left(\frac{1}{2}\right)\) . . . "the angle whose sine is \(\displaystyle \frac{1}{2}\)"

And that angle is \(\displaystyle \L\,\frac{\pi}{6}\) . . . right?