# [SOLVED] inverse of angles

• April 2nd 2009, 01:01 PM
lsnyder
[SOLVED] inverse of angles
Hey everyone,
I have a test tomorrow and having some difficulty with inverses. so if anyone can help, I'd really appreciate it. Also, I am a visual learner so if it is shown I understand a lot more and can able to apply the same principles I have seen to other problems.

So can someone show me how to do these:

1) Find the inverse function for

2) Evaluate each of the inverse trig functions in radians rounded to 4 decimal places

https://webwork.uncc.edu/webwork2_fi...0522a264c1.png https://webwork.uncc.edu/webwork2_fi...ddf16c4091.png
• April 3rd 2009, 11:31 AM
Hello lsnyder
Quote:

Originally Posted by lsnyder
1) Find the inverse function for

$0\le x\le\frac{\pi}{14}\Rightarrow 0\le 7x\le \frac{\pi}{2}$ and $f(x)$ is therefore is a one-to-one function, with values in the range $0\le f(x) \le 1$. So we can find its inverse, using $\arccos$, taking the angle in the range $0$ to $\frac{\pi}{2}$.

So, let $y = f(x) = \cos(7x)$

Then $\arccos(y) = 7x$

$\Rightarrow x = \tfrac17\arccos(y)$

So the inverse function (replacing $y$ by $x$) is $f^{-1}(x) = \tfrac17\arccos(x)$

Quote:

2) Evaluate each of the inverse trig functions in radians rounded to 4 decimal places

https://webwork.uncc.edu/webwork2_fi...0522a264c1.png https://webwork.uncc.edu/webwork2_fi...ddf16c4091.png
I assume that you have a calculator that will give you inverse tangents and inverse sines. Then:

$x = \text{arccot} (1.5)$

$\Rightarrow\cot(x) = 1.5$

$\Rightarrow \tan(x) = \frac{1}{1.5} = \frac23$

$\Rightarrow x = \arctan\Big(\frac23 \Big) = 0.5880$ to 4 d.p.

And $x = \csc^{-1}(8)$

$\Rightarrow \csc(x) = 8$

$\Rightarrow \sin(x) = \tfrac18$

$x = 0.1253$ to 4 d.p.