# Stumped on a trig simplification

• August 31st 2008, 07:08 PM
leftyguitarjoe
Stumped on a trig simplification
Some coaching on this would be nice.

All I'm working with is a formula sheet with some identities in it.

Any help at all would be greatly appreciated.

Simplify the expression.
cos(2tan-1(x))
• August 31st 2008, 07:19 PM
Chris L T521
Quote:

Originally Posted by leftyguitarjoe
Some coaching on this would be nice.

All I'm working with is a formula sheet with some identities in it.

Any help at all would be greatly appreciated.

Simplify the expression.
cos(2tan-1(x))

Note that the trig expression takes on the form $\cos(2\vartheta)$, where $\vartheta=\tan^{-1}x$

Recall that $\cos(2\vartheta)=1-2\sin^2(\vartheta)$

Now let us set up a triangle that represents the angle $\vartheta=\tan^{-1}x$.

We see that the opposite leg has a value of x, and the adjacent has a value of 1. Thus, the value of the hypotenuse can be found after applying the Pythagorean formula:

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

Now, let us evaluate the expression:

$1-2\sin^2(\vartheta)=1-2(\sin\vartheta)^2$

$\sin\vartheta$ can be determined from the triangle we constructed.

Thus, $\sin\vartheta=\frac{opp}{hyp}=\frac{x}{\sqrt{1+x^2 }}$

Substitute this into the equation and simplify:

$1-2(\sin\vartheta)^2=1-2\left(\frac{x}{\sqrt{1+x^2}}\right)^2=\dots$

I'll leave the rest for you to do.

I hope this makes sense! (Sun)

--Chris
• August 31st 2008, 07:25 PM
Jhevon
Quote:

Originally Posted by Chris L T521
Note that the trig expression takes on the form $\cos(2\vartheta)$, where $\vartheta=\tan^{-1}x$

Recall that $\cos(2\vartheta)=1-2\sin^2(\vartheta)$

Now let us set up a triangle that represents the angle $\vartheta=\tan^{-1}x$.

We see that the opposite leg has a value of x, and the adjacent has a value of 1. Thus, the value of the hypotenuse can be found after applying the Pythagorean formula:

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

Now, let us evaluate the expression:

$1-2\sin^2(\vartheta)=1-2(\sin\vartheta)^2$

$\sin\vartheta$ can be determined from the triangle we constructed.

Thus, $\sin\vartheta=\frac{opp}{hyp}=\frac{x}{\sqrt{1+x^2 }}$

Substitute this into the equation and simplify:

$1-2(\sin\vartheta)^2=1-2\left(\frac{x}{\sqrt{1+x^2}}\right)^2=\dots$

I'll leave the rest for you to do.

I hope this makes sense! (Sun)

--Chris

what's with the $\vartheta$ ? why can't you use $\theta$ like everyone else? (Tongueout)
• August 31st 2008, 07:26 PM
leftyguitarjoe
oh, I made a typo that might affect something

cos(tan^-1 (x))
• August 31st 2008, 07:28 PM
Chris L T521
Quote:

Originally Posted by Jhevon
what's with the $\vartheta$ ? why can't you use $\theta$ like everyone else? (Tongueout)

I'm sorry that I'm different!! (Worried)

Oh wait...you already knew that (Rofl)

Plus, $\vartheta$ looks cooler. $\theta$ is too boring :p

--Chris
• August 31st 2008, 07:28 PM
Chris L T521
Quote:

Originally Posted by leftyguitarjoe
oh, I made a typo that might affect something

cos(tan^-1 (x))

That does effect it...

--Chris
• August 31st 2008, 07:29 PM
Chop Suey
Quote:

Originally Posted by leftyguitarjoe
oh, I made a typo that might affect something

cos(tan^-1 (x))

You can still do it on your own. Use the answers given for the typo question as a reference. :)
• August 31st 2008, 07:31 PM
Chris L T521
Quote:

Originally Posted by Chris L T521
Note that the trig expression takes on the form $\cos(2\vartheta)$, where $\vartheta=\tan^{-1}x$

Recall that $\cos(2\vartheta)=1-2\sin^2(\vartheta)$

Now let us set up a triangle that represents the angle $\vartheta=\tan^{-1}x$.

We see that the opposite leg has a value of x, and the adjacent has a value of 1. Thus, the value of the hypotenuse can be found after applying the Pythagorean formula:

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

Now, let us evaluate the expression:

$1-2\sin^2(\vartheta)=1-2(\sin\vartheta)^2$

$\sin\vartheta$ can be determined from the triangle we constructed.

Thus, $\sin\vartheta=\frac{opp}{hyp}=\frac{x}{\sqrt{1+x^2 }}$

Substitute this into the equation and simplify:

$1-2(\sin\vartheta)^2=1-2\left(\frac{x}{\sqrt{1+x^2}}\right)^2=\dots$

I'll leave the rest for you to do.

I hope this makes sense! (Sun)

--Chris

It doesn't change much:

Keep this bit:

Quote:

Now let us set up a triangle that represents the angle $\vartheta=\tan^{-1}x$.

We see that the opposite leg has a value of x, and the adjacent has a value of 1. Thus, the value of the hypotenuse can be found after applying the Pythagorean formula:

$x^2+1=h^2\implies h=\sqrt{1+x^2}$
But just evaluate $\cos\vartheta$

$\cos\vartheta=\dots$

It should be obvious if you constructed the triangle properly

--Chris
• August 31st 2008, 07:32 PM
leftyguitarjoe
Quote:

Originally Posted by Chop Suey
You can still do it on your own. Use the answers given for the typo question as a reference. :)

aye gotchya(Wink)