Trigonometric Functions

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math

Are real function which relate an angle of a right angled triangle to the ratios of the 2 sides of that triangle.

Sine is classically a trigonometric function which specifies how the height of a circle changes as we move a hypothetical angle across it anti-clockwise from the top-right quadrant. At \( \theta = 360 = 0, \; \sin \theta = 0 \), at \( \theta = 90, \; \sin \theta = 1 \), at \( \theta = 180, \; \sin \theta = 0 \), at \( \theta = 270, \; \sin \theta = -1 \).

Cosine is the same as Sine except its \( 90^{\degrees} \) out of phase with it: \( \cos x = \sin (x + 90) \).

Tangent is the ratio of the sine of an angle to its cosine, that is to say \( \tan x = \frac{\sin x}{\cos x} \).

We define the Cotangent (\( \cot x \)), Secant (\( \sec x \)) and Cosecant (\( \cosec x \)) functions to be the reciprocals (1 over) the Tangent, Cosine and Cosecant functions respectively.

We also define the inverse trigonometric functions \( \sin^{-1}, \cos^{-1}, tan^{-1} \) as the inverse functions for \( \sin, \cos, \text{and} \tan \) respectively. Each of these reverse the affect of their matching function, that is to say \( \sin^{-1} (\sin x) = x \). These functions are commonly referred to as the arc functions, for example \( \sin^{-1} \) is the arc-sine function.

Hint: If you have trouble associating the 1 over variants with the originals, remember both of them have the same character at the 3rd index of their name.

Identities

Pythagorean Theorems

Pythagoras's theorem states: \[ \sin^2 x + \cos^2 x \equiv 1 \]

Using the definitions of secant, cosecant and cotangent we can adapt this into the variant theorems:

\begin{align} \label{eq:pythagorean-theorem} \sin^2 x + \cos ^2 x &\equiv 1 \
1 + \tan^2 x &\equiv \sec^2 x \
1 + \cot^2 x &\equiv \cosec^2 x \end{align}

Professor Terror

\begin{align} \sin(A \pm B) &\equiv \sin A \cos B \pm \cos A \sin B \
\label{eq:terror-cos} \cos(A \pm B) &\equiv \cos A \cos B \mp \sin A \sin B \
\tan(A \pm B) &\equiv \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}, \qquad \text{Where $A \pm B \neq (k + \frac{1}{2} \pi)$} \end{align}

\begin{align} \sin A + \sin B &\equiv 2 \sin \frac{A + B}{2} \cos \frac{A - B}{2} \
\sin A - \sin B &\equiv 2 \cos \frac{A + B}{2} \sin \frac{A - B}{2} \label{eq:terror-sin-amb} \
\cos A + \cos B &\equiv 2 \cos \frac{A + B}{2} \cos \frac{A - B}{2} \
\cos A - \cos B &\equiv -2 \sin \frac{A + B}{2} \sin \frac{A - B}{2} \end{align}

Double Angle

\begin{align} \sin 2x &\equiv 2 \sin x \cos x \
\cos 2x &\equiv \cos^2 x - \sin^2 x \
\cos 2x &\equiv 1 - 2 \sin^2 x \
\cos 2x &\equiv 2 \cos^2 x - 1 \
\tan 2x &\equiv \frac{2 \tan x}{1 - \tan^2 x} \end{align}

Misc

\begin{align*} \cos (-B) &= \cos B \
\sin (-B) &= - \sin B \end{align*}

Cosine rule

\begin{align} a^2 = b^2 + c^2 - 2bc \cos{A} \end{align}

Applications of Trigonometric Identities

When asked to solve a product of sines or cosines with weird angles, it helps to be able to redefine it as a summation.

For example consider the problem \[ \int \cos 3x \sin 2x \,dx \] Using the identity eq:terror-sin-amb we can define two simultaneous equations \( \frac{A+B}{2} = 3x \) and \( \frac{A-B}{2} = 2x \). The summation of these two equations gives \( A = 5x \) and the difference gives \( B = x \). We can substitute these back into the identity (accounting for the missing multiplier) to get \[ \cos 3x \sin 2x = \frac{1}{2} [\sin 5x - \sin x] \] This is a considerably simpler integration to solve than what we had originally.