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: Recall that $\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}$
So: $\displaystyle \tan \left(\arcsin v\right) = \frac{\sin \left(\arcsin v\right)}{\cos \left(\arcsin v\right)} = \frac{v}{\sqrt{1 - \big[\sin \left(\arcsin v\right)\big]^2}} = ...$
#2: Similar to what was done in the denominator of #1, recall that $\displaystyle \cos^2 x = 1 - \sin^2x$
So: $\displaystyle \sin \left(\arccos v\right) = \sqrt{1 - \big[\cos \left(\arccos v\right)\big]^2}$
#1: If we were to plot a triangle that would represent $\displaystyle \sin^{-1}v$, 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:
$\displaystyle v^2+b^2=1\implies b=\sqrt{1-v^2}$
Now, since $\displaystyle \tan\vartheta=\frac{opp}{adj}$, and $\displaystyle opp=v~and~adj=\sqrt{1-v^2}$ we see then that $\displaystyle \tan\vartheta=\tan\left(\sin^{-1}v\right)=\frac{v}{\sqrt{1-v^2}}$
$\displaystyle \mathbb{Q.E.D.}$
#2: If we were to plot a triangle that would represent $\displaystyle \cos^{-1}v$, 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:
$\displaystyle v^2+b^2=1\implies b=\sqrt{1-v^2}$
Now, since $\displaystyle \sin\vartheta=\frac{opp}{hyp}$, and $\displaystyle opp=\sqrt{1-v^2}~and~hyp=1$ we see then that $\displaystyle \sin\vartheta=\sin\left(\cos^{-1}v\right)=\sqrt{1-v^2}$
$\displaystyle \mathbb{Q.E.D.}$
Does this make sense??
--Chris
Hello, magentarita!
Another approach . . .
Recall that an inverse trig expression is an angle.Establishing an identity involving inverse trigonometric functions.
(1) Show that: .$\displaystyle \tan[\arcsin(v)] \:= \:\frac{v}{\sqrt{1 - v^2}}$
We have: .$\displaystyle \tan\underbrace{[\arcsin(v)]}_{\text{some angle }\theta}$
That is: .$\displaystyle \theta$ is an angle whose sine is $\displaystyle v.$
We have: .$\displaystyle \sin\theta \:=\:\frac{v}{1} \:=\:\frac{opp}{hyp}$
$\displaystyle \theta$ is in a right triangle with: .$\displaystyle opp = v,\;hyp = 1$
. . and we want: .$\displaystyle \tan\theta$Using Pythagorus, we find that .$\displaystyle adj = \sqrt{1-v^2}$Code:* 1 * | * | v * θ | * - - - - - - - * ____ √1-vē
Therefore: .$\displaystyle \tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{v}{\sqrt{1-v^2}} $
Ahh, Chris beat me to it . . .
.