By doing some right triangle gymnastics, we can derive things like cos &#x2061;<!-- ⁡ -->

tapiat0wa4c

tapiat0wa4c

Answered question

2022-06-04

By doing some right triangle gymnastics, we can derive things like
cos ( arctan x ) = 1 1 + x 2 , for x > 0 cos ( arcsin x ) = 1 x 2
cos ( arcsin x ) = 1 x 2
tan ( arcsin x ) = x 1 x 2
What about arctan cos ( x ), arcsin ( tan x ), etc?

Answer & Explanation

treskjeqlalo

treskjeqlalo

Beginner2022-06-05Added 3 answers

The identities you list can all be derived by expressing the trigonometric functions in terms of each other, e.g. (glossing over sign issues)
tan x = sin x cos x = sin x 1 sin 2 x ,
and thus
tan ( arcsin x ) = sin ( arcsin x ) 1 sin 2 ( arcsin x ) = x 1 x 2 .
So these identities hold because there are identities between the trigonometric functions, which in a sense are due to Euler's formula.
You can get similar identities in the other direction for arcsin and arccos because there's a suitable relationship between these two:
arccos x = π 2 arcsin x ,
and thus
arccos ( sin x ) = π 2 arcsin ( sin x ) = π 2 x .
That you can't get them for arctan cos ( x ) and arcsin ( tan x ) has to do with the fact that the identities between the corresponding inverse trigonometric functions have different arguments. For instance,
arctan x = arcsin x x 2 + 1 ,
but you can't turn this into a formula for arctan ( sin x ) because the argument on the right-hand side isn't x.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?