# Trigonometric Functions and Inverse Trigonometric Functions

• Feb 19th 2010, 07:03 AM
NOX Andrew
Trigonometric Functions and Inverse Trigonometric Functions
How would you derive the relationships between the trigonometric functions and the inverse trigonometric functions, such as:

$\sin(\arccos(x)) = \sqrt{1 - x^2}$

$\sin(\arctan(x)) = \frac{x}{\sqrt{1 + x^2}}$
• Feb 19th 2010, 10:32 AM
running-gag
Quote:

Originally Posted by NOX Andrew
How would you derive the relationships between the trigonometric functions and the inverse trigonometric functions, such as:

$\sin(\arccos(x)) = \sqrt{1 - x^2}$

$\sin(\arctan(x)) = \frac{x}{\sqrt{1 + x^2}}$

Hi

Using $\cos^2(\arccos(x))+\sin^2(\arccos(x)) = 1$ and $\cos(\arccos(x)) = x$

it comes $x^2+\sin^2(\arccos(x)) = 1$
• Feb 19th 2010, 10:37 AM
running-gag
Quote:

Originally Posted by NOX Andrew
How would you derive the relationships between the trigonometric functions and the inverse trigonometric functions, such as:

$\sin(\arccos(x)) = \sqrt{1 - x^2}$

$\sin(\arctan(x)) = \frac{x}{\sqrt{1 + x^2}}$

Use $\sin^2 \alpha = \frac{\tan^2 \alpha}{1+\tan^2 \alpha}$ applied to $\alpha = \arctan(x)$