### Trigonometry

• Utilize trigonometric identities to verify other identities.
• Be able to graph trigonometric functions.
• Apply trigonometric ratios and identities to solve problems related to distance and angle calculations.

## Basic Trigonometric Definitions

The three basic trigonometric ratios are defined as follows:

$\cos\ \theta = \dfrac{\mathrm{Adjacent}}{\mathrm{Hypotenuse}} = \dfrac{x}{r}$ $\sin\ \theta = \dfrac{\mathrm{Opposite}}{\mathrm{Hypotenuse}} = \dfrac{y}{r}$ $\tan\ \theta = \dfrac{\mathrm{Opposite}}{\mathrm{Adjacent}} = \dfrac{x}{y}$

The remaining trigonometric functions can be defined as reciprocals: \begin{aligned} \cot\ \theta &= \dfrac{1}{\tan\ \theta}\\ \sec\ \theta &= \dfrac{1}{\cos\ \theta}\\ \csc\ \theta &= \dfrac{1}{\sin\ \theta}\\\end{aligned}

## Trigonometric Ratios of Negative Angles:

\begin{aligned} \sin\ (-\theta) &= -\sin\ \theta\\ \cos\ (-\theta) &=\ \ \cos\ \theta\\ \tan\ (-\theta) &= -\tan\ \theta\\ \end{aligned}

## Fundamental Trigonometric Identities:

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

\begin{aligned} \cos(\alpha + \beta) &= \cos\ \alpha \cos\ \beta - \sin\ \alpha \sin\ \beta\\ \sin(\alpha + \beta) &= \sin\ \alpha \cos\ \beta + \cos\ \alpha \sin\ \beta \end{aligned}

## Trigonometric Formulas for Multiple Angles

$\cos\ 2\alpha = \cos^2 \alpha - \sin^2 \alpha$ Replacing $\cos^2 \alpha$ with $1-\sin^2 \alpha$ or replacing $\sin^2 \alpha$ with $1-\cos^2 \alpha$, we get two additional formulae for $\cos\ 2\alpha$

\begin{aligned} \cos\ 2\alpha &= 1 - 2\ \sin^2 \alpha \\ \cos\ 2\alpha &= 2\ \cos^2 \alpha - 1\end{aligned} $\sin\ 2\alpha = 2\ \sin \alpha \cos \alpha$

$\tan\ 2\alpha = \dfrac{2\tan\ \alpha}{1-\tan^2\ \alpha}$

### Solved Example:

#### TRIG-01

If x and y are complementary angles, then:

Solution:
Two angles are called complementary if their measures add to 90 degrees. $x + y = 90$ or, $y = 90 - x$ $\cos y = \cos (90 - x) = \sin x$

### Solved Example:

#### TRIG-02

If $\sin \theta + \sin^2 \theta = 1$, then $\cos^2 \theta + \cos^4 \theta =$

Solution:

\begin{aligned} \sin \theta + \sin^2 \theta &= 1\\ \sin \theta &= 1 - \sin^2 \theta = \cos^2 \theta ........ \mathrm{eqn(A)} \end{aligned}

The given expression, \begin{aligned} \cos^2 \theta + \cos^4 \theta &= \cos^2 \theta ( 1 + \cos^2 \theta)\\ &= \sin \theta ( 1+ \sin \theta ) ... (\mathrm{From\ eqn(A)\ } \cos^2 \theta= \sin \theta)\\ &= \sin \theta+ \sin^2 \theta \\ &= 1 ........ (\mathrm{Given\ identity})\end{aligned}