#1: Recall that
So:
#2: Similar to what was done in the denominator of #1, recall that
So:
#1: If we were to plot a triangle that would represent , 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:
Now, since , and we see then that
#2: If we were to plot a triangle that would represent , 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:
Now, since , and we see then that
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: .
We have: .
That is: . is an angle whose sine is
We have: .
is in a right triangle with: .
. . and we want: .Using Pythagorus, we find that .Code:* 1 * | * | v * θ | * - - - - - - - * ____ √1-vē
Therefore: .
Ahh, Chris beat me to it . . .
.