The trigonometric functions relate the angles in a right triangle to the ratios of the sides. Given a triangle like this:

The basic trigonometric functions would be defined, for \( 0^\circ < \theta < 90^{\circ} \) as:

\[\begin{array}{lcr} \sin \theta = \frac{a}{b} & \cos \theta = \frac{c}{b} & \tan \theta = \frac{a}{c} \end{array} \]

However, a more useful definition comes from the "unit circle." If we describe a circle with a radius of 1 unit, centered at the origin, then the angle \( \theta \) inside the circle describes a right triangle when when we drop a perpendicular to the \( x \)-axis from the point of intersection with the circle.

Notice that the right triangle so described has a hypotenuse equal to the radius of the circle, an adjacent side length of \( x \), and an opposite side length of \( y \). This gives rise naturally to the following refined definitions:

\[\begin{array}{lcr} \sin \theta = \frac{y}{r} & \cos \theta = \frac{x}{r} & \tan \theta = \frac{y}{x} \end{array} \]

In the unit circle, as shown, since the radius is 1, this simplifies to \( \sin \theta = y \), etc.

These definitions have the advantage of being compatible with the triangle definition above as well as allowing the evaluation of angles corresponding to any real number.

There are certain values of these functions which are useful to remember. They are:

\[ \begin{array} {| c | c | c | c | c | c |} \hline \theta & 0^\circ & 30^\circ & 45^\circ & 60^\circ & 90^\circ\\ \hline \sin \theta & \frac {\sqrt{0}} {2} & \frac {\sqrt{1}} {2} & \frac {\sqrt{2}} {2} & \frac {\sqrt{3}} {2} & \frac {\sqrt{4}} {2} \\ \hline \cos \theta & \frac {\sqrt{4}} {2} & \frac {\sqrt{3}} {2} & \frac {\sqrt{2}} {2} & \frac {\sqrt{1}} {2} & \frac {\sqrt{0}} {2}\\ \hline \tan \theta & 0 & \frac { 1}{\sqrt{3} } & 1 & \sqrt{3} & \mbox{undefined} \\ \hline \end{array}\]

The reason for writing them in this way, is to aid remembering these terms. For example, the numerator for \( \sin \theta\) is simply the square root of 0, 1, 2, 3, 4.

## Evaluate \( \sin 225^\circ \) exactly.

Notice that the problem reduces to finding the value of \( y \) in the above picture.

Since \( 225^\circ-180^\circ = 45^\circ \), \( \theta \) makes an angle of \( 45^\circ \) with the negative \( x \)-axis. Further, since this is a right triangle, the angles must be \( 90^\circ + 45^\circ + 45^\circ = 180^\circ \), which means it is isosceles. Therefore, \( | x | = |y| \).

By the Pythagorean Theorem, \[ \begin{align} x^2 + y^2 &= 1 \\ 2y^2 &= 1 \\ y^2 &= \frac{1}{2} \\ y &= \pm \sqrt{\frac{1}{2}}=\pm \frac{\sqrt{2}}{2} \end{align} \]

By inspection, \( y \) is negative, so \( \sin 225^\circ = -\frac{\sqrt{2}}{2} \).

Note:most values for \( \theta \) will be difficult or impossible to evaluate exactly in this way, so we often use a calculator to evaluate the approximate value of the function (see the table above for more exact values).

## Surveying

A surveyor in a helicopter at an elevation of 1000 meters measures the angle of depression to the far edge of an island as \( 24^\circ \) and the angle of depression to the near edge is \( 31^\circ \). How wide is the island, to the nearest meter?

SolutionLet the horizontal distance between the helicopter and the island be \( d \), and the width of the island be \( w \). Then \( \tan 24^\circ = \frac{1000}{d+w} \) and \( \tan 31^\circ = \frac{1000}{d} \). Thus \( d = \frac{1000}{\tan 31^\circ} \). Substituting:

\[ \begin{align} \tan 24^\circ &= \frac{1000}{\frac{1000}{\tan 31^\circ}+w} \\ \tan 24^\circ \left(\frac{1000}{\tan 31^\circ}+w \right) &= 1000 \\ w &= \frac{1000\left( 1 - \frac{\tan 24^\circ}{ \tan 31^\circ }\right)}{\tan 24^\circ} \\ w & \approx 581.75 \end{align} \]

Thus the width of the island is 582 meters.

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