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pre-calculus

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basic trigonometry

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1.2

Basic Trigonometry

Trigonometric Values Using Points

Some questions will provide you with a point on the terminal side of an angle, and you will be expected to find the exact values of each of the six trigonometric functions with this information.

Recall the meaning of each trignonometric function in terms of a triangle. See the following table for these definitions.

\( \sin \) \(\frac{\text{opposite}}{\text{hypotenuse}}\)
 \( \cos \) \(\frac{\text{adjacent}}{
\text{hypotenuse}
}\)
\( \tan \) \(\frac{\text{opposite}}{
\text{adjacent}
}\)		 \( \sec \) \(\frac{\text{hypotenuse}}{
\text{adjacent}
}\)
\( \csc \) \(\frac{\text{hypotenuse}}{
\text{opposite}
}\)		 \( \cot \) \(\frac{\text{adjacent}}{
\text{opposite}
}\)

The Pythagorean Theorem, sometimes known as the distance formula, can determine the distance between two points in a Cartesian plane or the hypotenuse of a triangle.

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

  1. If \( (-7,24) \) is a point on the terminal side of an angle A, find the exact values of each trignonometric function.
    2. First, let's find the hypotenuse of the "triangle" formed by the given point. \[\begin{align} x^2+y^2&=r^2 \\ (-7)^2+24^2&=r^2 \\ r&=25 \end{align}\] Next, we can simply plug in the three numbers we now have (\( -7,24,25 \)) to the trigonometric functions. \[\begin{align} \sin A&=\frac{24}{25} \\ \cos A&=\frac{-7}{25} \\ \tan A&=\frac{-24}{7} \\ \sec A&=\frac{-25}{7} \\ \csc A&=\frac{25}{24} \\ \cot A&=\frac{-7}{24} \end{align}\]
  2. If \( \cot B=\frac{\sqrt{11}}{5} \), and B is in Quadrant III, find the exact values of each of the following.
    3. We know that \(\cot\) is \( \frac{\text{adjacent}}{\text{opposite}} \), so we can use the values given above in the distance formula to find the hypotenuse. \[\begin{align} x^2+y^2&=r^2 \\ \sqrt{11}^2+5^2&=r^2 \\ r&=6 \end{align}\] Next, we can simply plug in the three numbers we now have (\( 5,6,\sqrt{11} \)) to the trigonometric functions. \[\begin{align} \sin A&=\frac{-5}{6} \\ \cos A&=\frac{\sqrt{11}}{6} \\ \tan A&=\frac{5}{\sqrt{11}} \\ \sec A&=\frac{-6}{\sqrt{11}} \\ \csc A&=\frac{-6}{5} \\ \cot A&=\frac{\sqrt{11}}{5} \end{align}\]

Unit Circles

A unit circle is a graphical tool used in trigonometry used to work with degrees, radians, and exact values. Unit circles are useful for understanding the relationship between trigonometry, triangles, and angles.

Source

Notice some key facts about the unit circle in the picture above:

  • The circle has a radius of exactly \(1\). You can see this from any of the points along the axes.
  • The angles are represented in both degrees and radians. Degrees are more commonly used in geometry while radians are used more in calculus, trigonometry, and physics.
  • There are various points highlighted around the circle. These points correspond with an angle, and they are used in various ways.
  1. What is \( \cos\frac{7\pi}{6} \)?
    2. \[ \frac{-\sqrt{3}}{2} \]
  2. What is \( \sec\frac{7\pi}{4} \)?
    3. \[ \sqrt{2} \]

Converting Radians and Degrees

It is often useful to convert between radians and degrees.

To convert to radians from degrees, multiply the degree value by \( \frac{\pi}{180} \).

To convert to degrees from radians, multiply the radian value by \( \frac{180}{\pi} \).

  1. Simplify this trigonometric expression: \( \tan^2\frac{\pi}{6}+\sin^2\frac{\pi}{6} \).
    2. \[\begin{align} \tan^2\frac{\pi}{6}+\sin^2\frac{\pi}{6}&=\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{2}\right)^2 \\ &=\frac{1}{3}+\frac{1}{4} \\ &=\frac{7}{12} \end{align}\] Recall that \( \tan^2\frac{\pi}{6} \) means \( (\tan\frac{\pi}{6})^2 \).