Trigonometric Functions and Inverse Trigonometric Functions

**NOX Andrew** · February 19th 2010, 07:03 AM

How would you derive the relationships between the trigonometric functions and the inverse trigonometric functions, such as:

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

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

**running-gag** · February 19th 2010, 10:32 AM

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

**running-gag** · February 19th 2010, 10:37 AM

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 $\alpha = \arctan(x)$

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