# Inverse Trigonometric Functions

• Aug 23rd 2008, 02:51 PM
magentarita
Inverse Trigonometric Functions
Establishing an Identity Involving Inverse Trigonometric Functions.

(1) Show that tan[arcsin(v)] = v/(sqrt{1 - v^2})

(2) Show that sin[arccos(v)] = sqrt{1 - v^2}
• Aug 23rd 2008, 03:00 PM
o_O
#1: Recall that $\tan \theta = \frac{\sin \theta}{\cos \theta}$

So: $\tan \left(\arcsin v\right) = \frac{\sin \left(\arcsin v\right)}{\cos \left(\arcsin v\right)} = \frac{v}{\sqrt{1 - \big[\sin \left(\arcsin v\right)\big]^2}} = ...$

#2: Similar to what was done in the denominator of #1, recall that $\cos^2 x = 1 - \sin^2x$

So: $\sin \left(\arccos v\right) = \sqrt{1 - \big[\cos \left(\arccos v\right)\big]^2}$
• Aug 23rd 2008, 03:02 PM
Chris L T521
Quote:

Originally Posted by magentarita
Establishing an Identity Involving Inverse Trigonometric Functions.

(1) Show that tan[arcsin(v)] = v/(sqrt{1 - v^2})

(2) Show that sin[arccos(v)] = sqrt{1 - v^2}

#1: If we were to plot a triangle that would represent $\sin^{-1}v$, the leg opposite the angle would have a value of v, and the hypotenuse would have a value of one.

To solve for the side adjacent to the angle, apply pythagoras' theorem:

$v^2+b^2=1\implies b=\sqrt{1-v^2}$

Now, since $\tan\vartheta=\frac{opp}{adj}$, and $opp=v~and~adj=\sqrt{1-v^2}$ we see then that $\tan\vartheta=\tan\left(\sin^{-1}v\right)=\frac{v}{\sqrt{1-v^2}}$

$\mathbb{Q.E.D.}$

#2: If we were to plot a triangle that would represent $\cos^{-1}v$, the leg adjacent to the angle would have a value of v, and the hypotenuse would have a value of one.

To solve for the side opposite to the angle, apply pythagoras' theorem:

$v^2+b^2=1\implies b=\sqrt{1-v^2}$

Now, since $\sin\vartheta=\frac{opp}{hyp}$, and $opp=\sqrt{1-v^2}~and~hyp=1$ we see then that $\sin\vartheta=\sin\left(\cos^{-1}v\right)=\sqrt{1-v^2}$

$\mathbb{Q.E.D.}$

Does this make sense??

--Chris
• Aug 23rd 2008, 04:39 PM
magentarita
Thanks
Thank you both. Of course, Chris' method is much easier to follow.
• Aug 24th 2008, 09:49 AM
Soroban
Hello, magentarita!

Another approach . . .

Quote:

Establishing an identity involving inverse trigonometric functions.

(1) Show that: . $\tan[\arcsin(v)] \:= \:\frac{v}{\sqrt{1 - v^2}}$

Recall that an inverse trig expression is an angle.

We have: . $\tan\underbrace{[\arcsin(v)]}_{\text{some angle }\theta}$

That is: . $\theta$ is an angle whose sine is $v.$

We have: . $\sin\theta \:=\:\frac{v}{1} \:=\:\frac{opp}{hyp}$

$\theta$ is in a right triangle with: . $opp = v,\;hyp = 1$
. . and we want: . $\tan\theta$
Code:

                      *             1    *  |               *      | v           * θ        |       * - - - - - - - *             ____             √1-vē
Using Pythagorus, we find that . $adj = \sqrt{1-v^2}$

Therefore: . $\tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{v}{\sqrt{1-v^2}}$

Ahh, Chris beat me to it . . .
.
• Aug 24th 2008, 03:49 PM
magentarita
soroban
Quote:

Originally Posted by Soroban
Hello, magentarita!

Another approach . . .

Recall that an inverse trig expression is an angle.

We have: . $\tan\underbrace{[\arcsin(v)]}_{\text{some angle }\theta}$

That is: . $\theta$ is an angle whose sine is $v.$

We have: . $\sin\theta \:=\:\frac{v}{1} \:=\:\frac{opp}{hyp}$

$\theta$ is in a right triangle with: . $opp = v,\;hyp = 1$
. . and we want: . $\tan\theta$
Code:

                      *             1    *  |               *      | v           * θ        |       * - - - - - - - *             ____             √1-vē
Using Pythagorus, we find that . $adj = \sqrt{1-v^2}$

Therefore: . $\tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{v}{\sqrt{1-v^2}}$

Ahh, Chris beat me to it . . .
.

Thank you Soroban.