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}

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- August 23rd 2008, 01:51 PMmagentaritaInverse 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} - August 23rd 2008, 02:00 PMo_O
#1: Recall that

So:

#2: Similar to what was done in the denominator of #1, recall that

So: - August 23rd 2008, 02:02 PMChris L T521
#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 - August 23rd 2008, 03:39 PMmagentaritaThanks
Thank you both. Of course, Chris' method is much easier to follow.

- August 24th 2008, 08:49 AMSoroban
Hello, magentarita!

Another approach . . .

Quote:

Establishing an identity involving inverse trigonometric functions.

(1) Show that: .

**angle**.

We have: .

That is: . is an angle whose sine is

We have: .

is in a right triangle with: .

. . and we want: .Code:`*`

1 * |

* | v

* θ |

* - - - - - - - *

____

√1-vē

Therefore: .

Ahh, Chris beat me to it . . .

. - August 24th 2008, 02:49 PMmagentaritasoroban