# Exact Values - Inverse Trig Expressions

• Dec 5th 2011, 07:59 PM
Diesal
Exact Values - Inverse Trig Expressions
Okay so this was explained horribly to me in class, so I have no clue what I am doing.

I am stuck on these problems:
Sec^-1(1/2)

Arccos(sin(π/6))

tan(Csc^-1(2^-1))

• Dec 5th 2011, 08:28 PM
Amer
Re: Exact Values - Inverse Trig Expressions
$\sec x = \frac{1}{\cos x}$
I think there is a mistake in the first one
since if there is x such that $\sec ^{-1} \frac{1}{2} = x \Rightarrow sec x = \frac{1}{2}$

so $\cos x = \frac{1}{\sec x } = 2$ and there is not any real number x satisfies that

to solve these things you should know that

$\sin \left(\frac{\pi}{6}\right) = \frac{1}{2} = \cos \left(\frac{\pi}{3} \right)$

$\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} = \sin \left(\frac{\pi}{3}\right)$

the last one also has a problem
since if $\csc ^{-1} 2^{-1}= x \Rightarrow \csc x = 2^{-1} = \frac{1}{2}$
and this you give $\sin x = 2$ same problem

second one if you know that sin (pi/6) equal 1/2
you have to know what is the angle where cos x = 1/2 ?

$\cos ^{-1} (sin \frac{\pi}{6} ) = \cos ^{-1} ( \frac{1}{2}) =$
• Dec 5th 2011, 08:44 PM
Diesal
Re: Exact Values - Inverse Trig Expressions
Again I think it's asking for an exact numerical value for Sec^-1(1/2) as well as the last one.

I'm not sure how to find exact values of inverses on the unit circle, other than take the first value which arccos or arcsin will be equal to.
• Dec 5th 2011, 09:12 PM
Amer
Re: Exact Values - Inverse Trig Expressions
for example you know that

$\sin \left( \frac{\pi}{6} \right) = \frac{1}{2}$

$\sin ^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{6}$ since the above

sin , cos is a bounded functions in real field $-1 \leq \sin (x),\cos(x) \leq 1$

and I know that $\sec x = \frac{1}{\cos x }$

so when they ask to find
$\sec ^{-1} \left(\frac{1}{2}\right)$

I begin to let it equal to x then taking the sec function to both sides

$\sec ^{-1} \left(\frac{1}{2}\right) = x$

$\sec \left( \sec ^{-1} \left(\frac{1}{2}\right) \right) = \sec (x)$

but sec is the inverse of sec^-1 and I know that
$f(f^{-1}(x)) = x$

so you will have on the left hand side

$\frac{1}{2} = \sec (x)$

I know that $\sec (x) = \frac{1}{\cos (x) }$

$\frac{1}{2} = \frac{1}{\cos (x) }$

$\cos (x) = 2$

and there is not any real number like that you can check this here

i means it is imaginary number not real