Even and Odd and Pythagorean Identities

(Trig 10 Professor Leonard - Precalc 83 )

Even and Odd Identities

What even means – symetric about the Y access (as output). Therefore opposite inputs (on the x axis, therefore, positive and negative) give you equal outputs. i.e.: \(f(-\theta) = f(\theta)\). Odd functions have symettry around the origin, so can rotate graph 180 degrees, so opposite inputs give you opposite outputs. I.e. \(f(-\theta) = - f(\theta)\). Remember x is \(cos(\theta),\) and y maps to \(sin(\theta)\) Look at positive and negative \( \frac{pi}{6}: \frac{pi}{6} = (\frac{\sqrt(3)}{2}, \frac{1}{2}), -pi/6 = (\frac{\sqrt(3)}{2}, \frac{1}{2})\)

Summary:

Even: cos, sec
Odd: sin, csc, tan, cot

Stands to reason because opposite angles are flipped accross x angles, and cos relates to y so it doesn’t change.

Even functions	Odd functions
\(cos(- \theta) = cos(\theta)\)	\(sin(- \theta) = - sin(\theta)\)
\(sec(-\theta) = sec(\theta)\)	\(csc( -\theta) = -csc(\theta)\)
	\(tan(-\theta) = -tan(\theta)\)
	\(cot(-\theta) = -cot(\theta)\)

Pythagorean Identities

Formula of circle:

\[ x^2 + y^2 = r^2 \]

Of course, on a unit circle, this works out to

\[ x^2 + y^2 = 1 \]

We also know that on a unit circle:

\[ cos(\theta) = x\space;\space sin(\theta) = y \]

Thus on a unit circle:

\[ (cos\space\theta)^2 + (sin\space\theta)^2 = 1 \]

Note can also square it here:

\[ cos^2\space\theta + sin^2\space\theta = 1 \]

Can also write it as follows:

\[ sin^2\space\theta +cos^2\space\theta = 1 \]

From this we can easily derive:

\[\begin{split} sin^2\space\theta = 1 - cos^2\space\theta \\ cos^2\space\theta = 1 - sin^2\space\theta \end{split}\]

Again:

\[ sin^2\theta +cos^2\theta = 1 \]

Can divide by \(sin^2\theta\) or \(cos^2\theta\):

First divide by \(sin^2\theta\)

\( \frac{sin^2\theta}{sin^2\theta} + \frac{cos^2\theta}{sin^2\theta} = \frac{1}{sin^2\theta} \\ So... \\ 1 + \frac{cos^2\theta}{sin^2\theta} = \frac{1}{sin^2\theta} \\ so \\ 1 + cot^2\theta = csc^2\theta \\ \\ or \\ cot^2\theta = csc^2\theta -1 \\ or \\ csc^2\theta - cot^2\theta = 1 \\ \)

Next divide by \(cos^2\theta\)

\[ \frac{sin^2\theta}{cos^2\theta} + \frac{cos^2\theta}{cos^2\theta} = \frac{1}{cos^2\theta} \]

So…

\( tan^2\theta + 1 = sec^2\theta \\ or \\ tan^2\theta + = sec^2\theta -1 \\ or \\ sec^2\theta - tan^2\theta = 1 \)

Three main ones to remember are below:

\[\begin{split} sin^2\space\theta +cos^2\space\theta = 1 \\ \space\\ cot^2\theta + 1 = csc^2\theta \\ \space\\ tan^2\theta + 1 = sec^2\theta \end{split}\]

The rest are corrolaries you can do as needed