Trigonometric identities

Trigonometric Identities are used to manipulate trigonometry equations in certain ways. Here is a list of them:

Even-Odd Identities

The functions $\sin(x)$ , $\tan(x)$ , and $\csc(x)$ are odd, while $cos(x)$ , $\cot(x)$ , and $\sec(x)$ are even. In other words, the six trigonometric functions satisfy the following equalities:

$\sin (-x) = -\sin (x)$
$\cos (-x) = \cos (x)$
$\tan (-x) = -\tan (x)$
$\sec (-x) = \sec (x)$
$\csc (-x) = -\csc (x)$
$\cot (-x) = -\cot (x)$

These are derived by the unit circle definitions of trigonometry. Making any angle negative is the same as reflecting it across the x-axis. This keeps its x-coordinate the same, but makes its y-coordinate negative. Thus, $\sin(-x) = -\sin(x)$ and $\cos(-x) = \cos(x)$ .

Pythagorean Identities

The Pythagorean identities state that

$\sin^2x + \cos^2x = 1$
$1 + \cot^2x = \csc^2x$
$\tan^2x + 1 = \sec^2x$

Using the ratio definition of trigonometry, we apply Pythagorean Theorem on our triangle above to get that $a^2 + b^2 = c^2$ . If we divide by $c^2$ , we get $\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1$ , which is just $\sin^2 x + \cos^2 x =1$ . Using the unit circle definition of trigonometry, it's extremely easy to see that $\sin^2 x+ \cos^2 x =1$ .

To derive the other two Pythagorean identities, divide by either $\sin^2 (x)$ or $\cos^2 (x)$ and substitute the respective trigonometry in place of the ratios.

Angle Addition Identities

The trigonometric angle addition identities state the following identities:

$\sin(x + y) = \sin (x) \cos (y) + \cos (x) \sin (y)$
$\cos(x + y) = \cos (x) \cos (y) - \sin (x) \sin (y)$
$\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)}$

There are many proofs of these identities. For the sake of brevity, we list only one here.

Euler's identity states that $e^{ix} = \cos (x) + i \sin(x)$ . We have that $\begin{align*} \cos (x+y) + i \sin (x+y) &= e^{i(x+y)} \\ &= e^{ix} \cdot e^{iy} \\ &= (\cos (x) + i \sin (x))(\cos (y) + i \sin (y)) \\ &= (\cos (x) \cos (y) - \sin (x) \sin(y)) + i(\sin (x) \cos(y) + \cos(x) \sin(y)) \end{align*}$ By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas.

To derive the tangent addition formula, we reduce tangent to sine and cosine, divide both numerator and denominator by $\cos (x) \cos (y)$ , and simplify. $\begin{align*} \tan (x+y) &= \frac{\sin (x+y)}{\cos (x+y)} \\ &= \frac{\sin (x) \cos(y) + \cos(x) \sin(y)}{\cos (x) \cos (y) - \sin (x) \sin(y)} \\ &= \frac{\frac{\sin(x)}{\cos(x)} + \frac{\sin(y)}{\cos(x)}}{1 - \frac{\sin (x) \sin(y)}{\cos (x) \cos(y)}} \\ &= \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan(y)} \end{align*}$ as desired.

Double-Angle Identities

The trigonometric double-angle identities are easily derived from the angle addition formulas by just letting $x = y$ . Doing so yields:

$\sin (2x) = 2\sin (x) \cos (x)$
$\cos (2x) = \cos^2 (x) - \sin^2 (x)$
$\tan (2x) = \frac{2\tan (x)}{1-\tan^2 (x)}$

Cosine Double-Angle

Here are two equally useful forms of the cosine double-angle identity. Both are derived via the Pythagorean identity on the cosine double-angle identity given above.

$\cos (2x) = 1 - 2 \sin^2 (x)$
$\cos (2x) = 2 \cos^2 (x) - 1$

In addition, the following identities are useful in integration and in deriving the half-angle identities. They are a simple rearrangement of the two above.

$\sin^2 (x) = \frac{1 - \cos (2x)}{2}$
$\cos^2 (x) = \frac{1 + \cos (2x)}{2}$

Half-Angle Identities

The trigonometric half-angle identities state the following equalities:

$\sin (\frac{x}{2}) = \pm \sqrt{\frac{1 - \cos (x)}{2}}$
$\cos (\frac{x}{2}) = \pm \sqrt{\frac{1 + \cos (x)}{2}}$
$\tan (\frac{x}{2}) = \pm \sqrt{\frac{1 - \cos (x)}{1+\cos \theta}} = \frac{\sin x}{1 + \cos (x)} = \frac{1-\cos (x)}{\sin (X)}$

The plus or minus is not saying that there are two answers, but that the sine of the angle depends on the quadrant that the angle is in.

Take the two expressions listed in the cosine double-angle section for $\sin^2 (x)$ and $\cos^2 (x)$ , and substitute $\frac{1}{2} x$ instead of $x$ . Taking the square root then yields the desired half-angle identities for sine and cosine.

Sum-to-Product Identities

${\sin \theta + \sin \gamma = 2 \sin \frac{\theta + \gamma}2 \cos \frac{\theta - \gamma}2}$
${\sin \theta - \sin \gamma = 2 \sin \frac{\theta - \gamma}2 \cos \frac{\theta + \gamma}2}$
${\cos \theta + \cos \gamma = 2 \cos \frac{\theta+\gamma}2 \cos \frac{\theta-\gamma}2}$
${\cos \theta - \cos \gamma = -2 \sin \frac{\theta+\gamma}2 \sin \frac{\theta-\gamma}2}$

Law of Sines

Main article: Law of Sines

The extended Law of Sines states

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

Law of Cosines

Main article: Law of Cosines

The Law of Cosines states

$a^2 = b^2 + c^2 - 2bc\cos A.$

Law of Tangents

Main article: Law of Tangents

The Law of Tangents states that if $A$ and $B$ are angles in a triangle opposite sides $a$ and $b$ respectively, then

$\frac{\tan{\left(\frac{A-B}{2}\right)}}{\tan{\left(\frac{A+B}{2}\right)}}=\frac{a-b}{a+b} .$

A further extension of the Law of Tangents states that if $A$ , $B$ , and $C$ are angles in a triangle, then $\tan(A)\cdot\tan(B)\cdot\tan(C)=\tan(A)+\tan(B)+\tan(C)$

Other Identities

$\sin(90-\theta) = \cos(\theta)$
$\cos(90-\theta)=\sin(\theta)$
$\tan(90-\theta)=\cot(\theta)$
$\sin(180-\theta) = \sin(\theta)$
$\cos(180-\theta) = -\cos(\theta)$
$\tan(180-\theta) = -\tan(\theta)$
$e^{i\theta} = \cos \theta + i\sin \theta$ (This is also written as $\text{cis }\theta$ )
$|1-e^{i\theta}|=2\sin\frac{\theta}{2}$
$\left(\tan\theta + \sec\theta\right)^2 = \frac{1 + \sin\theta}{1 - \sin\theta}$
$\sin(\theta) = \cos(\theta)\tan(\theta)$
$\cos(\theta) = \frac{\sin(\theta)}{\tan(\theta)}$
$\sec(\theta) = \frac{\tan(\theta)}{\sin(\theta)}$
$\arctan(x) + \arctan(y) = \arctan \left( \dfrac{x+y}{1-xy} \right)$
$\sin^2(\theta) + \cos^2(\theta) + \tan^2(\theta) = \sec^2(\theta)$
$\sin^2(\theta) + \cos^2(\theta) + \cot^2(\theta) = \csc^2(\theta)$