# Thread: Solving inverse trig equations

1. ## Solving inverse trig equations

Hey, I have two problems here that I really don't know how to start. Hopefully someone can help me out!
Question
Prove that $\displaystyle cos(sin^{-1}x) = \sqrt{1-x^2}$

Solution
I don't know where to start with this problem. I know that $\displaystyle sin^{-1}x = y -> siny = x$. I don't know how to apply this to this equation. A similar problem in the examples of my text is: "simplify: $\displaystyle cos(tan^{-1}x)$.
To simplify this the text first let y = $\displaystyle tan^{-1}x$. Then it jumps to $\displaystyle sec^2y = 1 + tan^2y = 1 + x^2$. I don't understand this step in their example and this may prove to be the problem I am having.

I hope I am making sense here! I have another problem, but I figure once I get help with this one I can force my way though the second. Please help!

2. Originally Posted by Kakariki
Hey, I have two problems here that I really don't know how to start. Hopefully someone can help me out!
Question
Prove that $\displaystyle cos(sin^{-1}x) = \sqrt{1-x^2}$

Solution
I don't know where to start with this problem. I know that $\displaystyle sin^{-1}x = y -> siny = x$. I don't know how to apply this to this equation. A similar problem in the examples of my text is: "simplify: $\displaystyle cos(tan^{-1}x)$.
To simplify this the text first let y = $\displaystyle tan^{-1}x$. Then it jumps to $\displaystyle sec^2y = 1 + tan^2y = 1 + x^2$. I don't understand this step in their example and this may prove to be the problem I am having.

I hope I am making sense here! I have another problem, but I figure once I get help with this one I can force my way though the second. Please help!
There have been several questions quite similar to this one posted in recent days. Have you reviewed the threads in this subforum?

3. Originally Posted by Kakariki
Hey, I have two problems here that I really don't know how to start. Hopefully someone can help me out!
Question
Prove that $\displaystyle cos(sin^{-1}x) = \sqrt{1-x^2}$

Solution
I don't know where to start with this problem. I know that $\displaystyle sin^{-1}x = y -> siny = x$. I don't know how to apply this to this equation. A similar problem in the examples of my text is: "simplify: $\displaystyle cos(tan^{-1}x)$.
To simplify this the text first let y = $\displaystyle tan^{-1}x$. Then it jumps to $\displaystyle sec^2y = 1 + tan^2y = 1 + x^2$. I don't understand this step in their example and this may prove to be the problem I am having.

I hope I am making sense here! I have another problem, but I figure once I get help with this one I can force my way though the second. Please help!
Here's another way to view it.........

You've gotten to

$\displaystyle Sin^{-1}x=y\Rightarrow\ x=Siny$

therefore $\displaystyle y$ is an angle (make it acute in a right-angled triangle) and

$\displaystyle Siny=\displaystyle\frac{opposite}{hypotenuse}$

hence this ratio is

$\displaystyle x=\displaystyle\frac{kx}{k}$

so the opposite is "kx" and the hypotenuse is "k"

Using Pythagoras' theorem....$\displaystyle (adjacent)^2+(kx)^2=k^2\Rightarrow\ adj=\sqrt{k^2-(kx)^2}=k\sqrt{1-x^2}$

You can work this through to calculate

$\displaystyle \displaystyle\ Cosy=\frac{adjacent}{hypotenuse}$

4. For the similar example, they've let $\displaystyle y = \tan^{-1}x$ which means $\displaystyle x = \tan{y}$.

$\displaystyle \sec^2{y} = 1 + \tan^2{y}$ is a well-known variation of the Pythagorean Identity. So

$\displaystyle \sec^2{y} = 1 + x^2$ because $\displaystyle x = \tan{y}$

$\displaystyle \sec{y} = \sqrt{1 + x^2}$

$\displaystyle \frac{1}{\cos{y}} = \sqrt{1 + x^2}$

$\displaystyle \cos{y} = \frac{1}{\sqrt{1 + x^2}}$

$\displaystyle \cos{(\tan^{-1}{x})} = \frac{1}{\sqrt{1 + x^2}}$ bceause $\displaystyle y = \tan^{-1}{x}$.