7. Composition of (Inverse) Trigonometric Values

Regular Values:

Angle $\theta$	sin($\theta$)	cos($\theta$)	tan($\theta$)
0	0	1	0
$\frac{\pi }{6}$	$\frac{1}{2}$	$\frac{1}{2}\sqrt{3}$	$\frac{1}{3}\sqrt{3}$
$\frac{\pi }{4}$	$\frac{1}{2}\sqrt{2}$	$\frac{1}{2}\sqrt{2}$	1
$\frac{\pi }{3}$	$\frac{1}{2}\sqrt{3}$	$\frac{1}{2}$	$\sqrt{3}$
$\frac{\pi }{2}$	1	0	1

Matching Compositions:

$\arcsin (\sin (x))=x$ and $\sin (\arcsin (x))=x$

$\arccos (\cos (x))=x$ and $\cos (\arccos (x))=x$

$\arctan (\tan (x))=x$ and $\tan (\arctan (x))=x$

Example:

$\arctan (\tan (10))=10-3\pi$

Non-matching Compositions:

(Method 1) Using ${\sin }^2(\theta )+{\cos }^2(\theta )=1$

$\sin (\arccos (x))=?$

${\sin }^2(\arccos (x))+{\cos }^2(\arccos (x))=1$ -⇒ Use the Pythagorean Identity

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

${\sin }^2(\arccos (x))=1-x^2$ -⇒ Take the square root.

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

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

Non-matching Compositions:

(Method 2) Using Triangles

$\cos (\arctan (x))=?$

$\theta =\arctan (x)$

$\tan (\theta )=x$

$\cos (\theta )=\frac{1}{\sqrt{1+x^2}}$