Consider the unit circle (radius $1$) centered at the origin. For an angle $x$ measured counterclockwise from the positive horizontal axis, the terminal side meets the circle at the point $P=(\cos x,\ \sin x)$. Thus $\cos x$ is the horizontal coordinate and $\sin x$ is the vertical coordinate of $P$. The remaining functions are ratios: $\tan x=\frac{\sin x}{\cos x}$, $\cot x=\frac{\cos x}{\sin x}$, $\sec x=\frac{1}{\cos x}$ and $\csc x=\frac{1}{\sin x}$. Because $P$ lies on the circle, $-1\le\sin x\le 1$ and $-1\le\cos x\le 1$. The signs depend on the quadrant of $P$: in quadrant I all functions are positive; in II only $\sin$ and $\csc$; in III only $\tan$ and $\cot$; in IV only $\cos$ and $\sec$.
An angle can be measured in degrees or in radians. A full turn equals $360^\circ$ or $2\pi$ radians, so $180^\circ=\pi$ radians. Conversions use $1^\circ=\frac{\pi}{180}$ rad and $1\text{ rad}=\frac{180^\circ}{\pi}$. To convert, multiply: degrees to radians by $\frac{\pi}{180}$, radians to degrees by $\frac{180}{\pi}$. Degrees split further into minutes and seconds (DMS): $1^\circ=60'$ and $1'=60''$, so $1^\circ=3600''$. To turn a decimal degree such as $12.35^\circ$ into DMS, keep the whole part as degrees, multiply the fractional part by $60$ to get minutes, then multiply the leftover fractional minutes by $60$ to get seconds.
The Pythagorean identity holds for every angle: $$\sin^2 x+\cos^2 x=1.$$ Dividing it by $\cos^2 x$ or by $\sin^2 x$ gives the companions $1+\tan^2 x=\sec^2 x$ and $1+\cot^2 x=\csc^2 x$. The reciprocal and quotient relations tie the six functions together: $\tan x=\frac{\sin x}{\cos x}$, $\cot x=\frac{\cos x}{\sin x}=\frac{1}{\tan x}$, $\sec x=\frac{1}{\cos x}$, $\csc x=\frac{1}{\sin x}$. Parity: $\cos$ and $\sec$ are even, $\sin(-x)=-\sin x$, $\tan(-x)=-\tan x$, $\cot(-x)=-\cot x$, $\csc(-x)=-\csc x$. These identities let one express any single function through another, up to a sign fixed by the quadrant.
For any angles $a$ and $b$: $$\sin(a\pm b)=\sin a\cos b\pm\cos a\sin b,$$ $$\cos(a\pm b)=\cos a\cos b\mp\sin a\sin b,$$ $$\tan(a\pm b)=\frac{\tan a\pm\tan b}{1\mp\tan a\tan b}.$$ Note the sign pattern: for $\cos$ the sign on the right is opposite to the one on the left, while for $\sin$ it matches. These formulas evaluate angles like $75^\circ=45^\circ+30^\circ$ or $15^\circ=45^\circ-30^\circ$ from known special values, and they are the source of the double-angle and product formulas below.
Setting $b=a$ in the addition formulas gives the double-angle formulas: $$\sin 2x=2\sin x\cos x,\qquad \tan 2x=\frac{2\tan x}{1-\tan^2 x},$$ $$\cos 2x=\cos^2 x-\sin^2 x=2\cos^2 x-1=1-2\sin^2 x.$$ The last three forms are equivalent through $\sin^2 x+\cos^2 x=1$. Solving them for the squares yields the power-reduction (half-angle) formulas $$\cos^2 x=\frac{1+\cos 2x}{2},\qquad \sin^2 x=\frac{1-\cos 2x}{2},$$ and, with $x\to\frac{x}{2}$, $\ \tan\frac{x}{2}=\frac{1-\cos x}{\sin x}=\frac{\sin x}{1+\cos x}$.
Sums of like functions factor into products: $$\sin a+\sin b=2\sin\frac{a+b}{2}\cos\frac{a-b}{2},$$ $$\sin a-\sin b=2\cos\frac{a+b}{2}\sin\frac{a-b}{2},$$ $$\cos a+\cos b=2\cos\frac{a+b}{2}\cos\frac{a-b}{2},$$ $$\cos a-\cos b=-2\sin\frac{a+b}{2}\sin\frac{a-b}{2}.$$ Reversing them turns products into sums: $$\sin a\cos b=\tfrac{1}{2}\big[\sin(a+b)+\sin(a-b)\big],$$ $$\cos a\cos b=\tfrac{1}{2}\big[\cos(a+b)+\cos(a-b)\big],$$ $$\sin a\sin b=\tfrac{1}{2}\big[\cos(a-b)-\cos(a+b)\big].$$ For example $\cos 7x+\cos 5x=2\cos 6x\cos x$.
The special (reference) angles are worth memorizing. For $\sin$: $\sin 0^\circ=0$, $\sin 30^\circ=\frac{1}{2}$, $\sin 45^\circ=\frac{\sqrt{2}}{2}$, $\sin 60^\circ=\frac{\sqrt{3}}{2}$, $\sin 90^\circ=1$. For $\cos$ the values run in reverse: $\cos 0^\circ=1$, $\cos 30^\circ=\frac{\sqrt{3}}{2}$, $\cos 45^\circ=\frac{\sqrt{2}}{2}$, $\cos 60^\circ=\frac{1}{2}$, $\cos 90^\circ=0$. Then $\tan=\frac{\sin}{\cos}$ gives $\tan 0^\circ=0$, $\tan 30^\circ=\frac{\sqrt{3}}{3}$, $\tan 45^\circ=1$, $\tan 60^\circ=\sqrt{3}$, while $\tan 90^\circ$ is undefined. In radians these are the angles $0,\ \frac{\pi}{6},\ \frac{\pi}{4},\ \frac{\pi}{3},\ \frac{\pi}{2}$. For instance $\sin\frac{\pi}{6}=\frac{1}{2}$ and $\cos\frac{\pi}{3}=\frac{1}{2}$.
Trigonometric functions are periodic: $\sin$ and $\cos$ have period $360^\circ$ ($2\pi$), while $\tan$ and $\cot$ have period $180^\circ$ ($\pi$). Hence $\sin(x+360^\circ k)=\sin x$ for any integer $k$; angles differing by a whole number of turns are coterminal and share all function values. To evaluate a large or negative angle, first reduce it modulo $360^\circ$ (or $180^\circ$ for $\tan$) into $[0^\circ,360^\circ)$. Reduction formulas then relate any angle to an acute reference angle, for example $\sin(180^\circ-x)=\sin x$, $\cos(180^\circ-x)=-\cos x$, $\sin(180^\circ+x)=-\sin x$, $\cos(360^\circ-x)=\cos x$, and the co-function shifts $\sin(90^\circ-x)=\cos x$, $\cos(90^\circ-x)=\sin x$, $\tan(90^\circ-x)=\cot x$.
To invert a trigonometric function one restricts its domain so it becomes one-to-one. Then $\arcsin$ (written $\sin^{-1}$) has domain $[-1,1]$ and range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$; $\arccos$ has domain $[-1,1]$ and range $[0,\pi]$; $\arctan$ has all real inputs and range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. By definition $\arcsin y$ is the unique angle in that range whose sine is $y$. For example $\arcsin\frac{\sqrt{3}}{2}=\frac{\pi}{3}$ because $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$ and $\frac{\pi}{3}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$; likewise $\arccos\left(-\frac{1}{2}\right)=\frac{2\pi}{3}$ and $\arctan 1=\frac{\pi}{4}$. Always pick the value inside the stated range, not just any angle with the right ratio.
Compute $\sin 920^\circ$. Reduce modulo $360^\circ$: $920^\circ-2\cdot 360^\circ=920^\circ-720^\circ=200^\circ$, so $\sin 920^\circ=\sin 200^\circ$. The angle $200^\circ$ lies in quadrant III, where sine is negative, and its reference angle is $200^\circ-180^\circ=20^\circ$. Using $\sin(180^\circ+x)=-\sin x$ with $x=20^\circ$ gives $\sin 200^\circ=-\sin 20^\circ$. Therefore $\sin 920^\circ=-\sin 20^\circ$. As a second case, $\sin(-2580^\circ)$: since $-2580^\circ+8\cdot 360^\circ=-2580^\circ+2880^\circ=300^\circ$, we get $\sin(-2580^\circ)=\sin 300^\circ=-\sin 60^\circ=-\frac{\sqrt{3}}{2}$.
Convert $\frac{5\pi}{12}$ radians to degrees, minutes and seconds. First to degrees: $\frac{5\pi}{12}\cdot\frac{180^\circ}{\pi}=\frac{5\cdot 180^\circ}{12}=\frac{900^\circ}{12}=75^\circ$, which is an exact whole number, so $\frac{5\pi}{12}\text{ rad}=75^\circ 0' 0''$. For a value that is not whole, say $1.2$ radians: $1.2\cdot\frac{180^\circ}{\pi}\approx 68.7549^\circ$. The whole part is $68^\circ$. The fractional part times $60$ is $0.7549\cdot 60\approx 45.29'$, so $45'$. The leftover $0.29'$ times $60$ is about $17''$. Hence $1.2\text{ rad}\approx 68^\circ 45' 17''$.
Simplify $\frac{\cos x}{\csc x}+\frac{\sin x}{\sec x}$ and then simplify $\sec x-\sin x\tan x$. For the first, rewrite the reciprocals: $\csc x=\frac{1}{\sin x}$ so $\frac{\cos x}{\csc x}=\cos x\sin x$, and $\sec x=\frac{1}{\cos x}$ so $\frac{\sin x}{\sec x}=\sin x\cos x$. Their sum is $2\sin x\cos x=\sin 2x$. For the second, write everything over $\cos x$: $\sec x-\sin x\tan x=\frac{1}{\cos x}-\frac{\sin^2 x}{\cos x}=\frac{1-\sin^2 x}{\cos x}=\frac{\cos^2 x}{\cos x}=\cos x$, using $1-\sin^2 x=\cos^2 x$.
Find the exact value of $\cos 15^\circ$. Write $15^\circ=45^\circ-30^\circ$ and apply $\cos(a-b)=\cos a\cos b+\sin a\sin b$: $$\cos 15^\circ=\cos 45^\circ\cos 30^\circ+\sin 45^\circ\sin 30^\circ.$$ Substituting the special values $\cos 45^\circ=\sin 45^\circ=\frac{\sqrt{2}}{2}$, $\cos 30^\circ=\frac{\sqrt{3}}{2}$, $\sin 30^\circ=\frac{1}{2}$: $$\cos 15^\circ=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\frac{\sqrt{6}+\sqrt{2}}{4}.$$ The same method with $\sin(a-b)$ gives $\sin 15^\circ=\frac{\sqrt{6}-\sqrt{2}}{4}$.
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