Trigonometric Function of Arbitrary Angle

Trigonometry for Angles in Every Quadrant

When a clock's minute hand moves from $12$ to $6$, it sweeps through $180^\circ$. One complete rotation sweeps through $360^\circ$.

Unit Circle

To understand trigonometric functions of arbitrary angles, we use the unit circle, which has radius $1$ and center $O(0,0)$.

The following details connect the angle to its coordinates:

• Angle $\theta$ is always measured from the positive $x$-axis
• Positive direction is counterclockwise
• Every point on the circle has coordinates $(x, y)$

Definitions:

Function Signs in Each Quadrant

As the point moves around the circle, the $x$ and $y$ coordinates can be positive or negative. The signs of those coordinates determine the signs of the trigonometric functions in each quadrant.

Signs in each quadrant:

You do not have to rely on a language-specific mnemonic. Read the signs from the coordinates instead: $\sin\theta$ has the sign of $y$, $\cos\theta$ has the sign of $x$, and $\tan\theta=y/x$ is positive exactly when $x$ and $y$ have the same sign.

Reference Angle

For an angle whose terminal side does not lie on an axis, the reference angle is the unique acute angle $0^\circ<\alpha<90^\circ$ between that terminal side and the nearest $x$-axis. Its acute-angle values give the magnitudes of the original angle's trigonometric values. The quadrant determines their signs. For a quadrantal angle, such as $90^\circ$ or $180^\circ$, read the coordinates directly from the unit circle instead.

How to determine the reference angle $\alpha$ for a nonquadrantal angle $0^\circ<\theta<360^\circ$:

• Quadrant I: $\alpha = \theta$
• Quadrant II: $\alpha = 180^\circ - \theta$
• Quadrant III: $\alpha = \theta - 180^\circ$
• Quadrant IV: $\alpha = 360^\circ - \theta$

Determining Trigonometric Values

Here are systematic steps to determine trigonometric function values:

1. Simplify the angle (if greater than $360^\circ$ or negative)
2. Determine the quadrant where the angle lies
3. Calculate the reference angle
4. Use the reference angle value with the appropriate sign for the quadrant

Angle in Quadrant Two

Problem: Determine $\sin 120^\circ$, $\cos 120^\circ$, and $\tan 120^\circ$

Solution:

• Angle $120^\circ$ lies in quadrant II (since $90^\circ < 120^\circ < 180^\circ$)
• Reference angle: $\alpha = 180^\circ - 120^\circ = 60^\circ$
• In quadrant II: $\sin(+), \cos(-), \tan(-)$

Using special angle values for $60^\circ$:

Angle in Quadrant Three

Problem: Determine trigonometric values for angle $240^\circ$

Solution:

• Angle $240^\circ$ lies in quadrant III (since $180^\circ < 240^\circ < 270^\circ$)
• Reference angle: $\alpha = 240^\circ - 180^\circ = 60^\circ$
• In quadrant III: $\sin(-), \cos(-), \tan(+)$

Using special angle values for $60^\circ$:

Angle in Quadrant Four

Problem: Determine trigonometric values for angle $300^\circ$

Solution:

• Angle $300^\circ$ lies in quadrant IV (since $270^\circ < 300^\circ < 360^\circ$)

• Reference angle: $\alpha = 360^\circ - 300^\circ = 60^\circ$

• In quadrant IV: $\sin(-), \cos(+), \tan(-)$

Using special angle values for $60^\circ$:

Handling Special Angles

Angles outside the first quadrant, negative angles, and angles beyond one turn all reduce to a reference angle. The reference angle keeps the values you already know for the first quadrant, and only the sign can change, so the quadrant decides that sign.

Negative Angles

A negative angle means you move clockwise from the starting point. Convert it first into an angle between $0^\circ$ and $360^\circ$ so the known values apply.

When the angle is negative, we move clockwise. Use the properties:

• $\sin(-\theta) = -\sin \theta$ (odd function)
• $\cos(-\theta) = \cos \theta$ (even function)
• $\tan(-\theta) = -\tan \theta$ (odd function)

Example: $\sin(-30^\circ) = -\sin 30^\circ = -\frac{1}{2}$

Angles Greater than One Full Rotation

Use the periodicity property. Subtract or add multiples of $360^\circ$ until the angle is in the range of $0^\circ$ to $360^\circ$.

Example:

• Subtract two complete turns:

  $$\begin{aligned} &750^\circ - 2(360^\circ) \\ &= 750^\circ - 720^\circ \\ &= 30^\circ \end{aligned}$$

• Therefore $\sin 750^\circ = \sin 30^\circ = \frac{1}{2}$

Exercises

Each problem asks for a trigonometric value at an angle outside the first quadrant. Reduce the angle to its reference angle first, then apply the correct sign.

1. Determine the values of $\sin 315^\circ$, $\cos 315^\circ$, and $\tan 315^\circ$.

2. Calculate $\sin(-60^\circ) + \cos 210^\circ - \tan(-135^\circ)$.

3. If $\sin \theta = \frac{3}{5}$ and $\theta$ is in quadrant II, determine $\cos \theta$ and $\tan \theta$.

4. Simplify $\sin 840^\circ \cdot \cos(-330^\circ)$.

5. A windmill rotates $1050^\circ$ from its initial position. If the initial position of the blade is on the positive $x$-axis, determine the coordinates of the blade tip on the unit circle after this rotation.

Answer Key

1. For angle $315^\circ$, we need to determine its quadrant first.

   Since $270^\circ < 315^\circ < 360^\circ$, the angle is in quadrant IV.

   The reference angle is $360^\circ - 315^\circ = 45^\circ$.

   $$\sin 315^\circ = -\sin 45^\circ = -\frac{\sqrt{2}}{2}$$

   $$\cos 315^\circ = +\cos 45^\circ = \frac{\sqrt{2}}{2}$$

   $$\tan 315^\circ = -\tan 45^\circ = -1$$

2. Calculate each term separately. For $\sin(-60^\circ)$, use the odd function property.

   For $\cos 210^\circ$, the angle is in quadrant III with reference $30^\circ$.

   For $\tan(-135^\circ)$, first convert to positive angle.

   $$\sin(-60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2}$$

   $$\cos 210^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2}$$

   $$\begin{aligned} \tan(-135^\circ) &= -\tan 135^\circ \\ &= -(-\tan 45^\circ) \\ &= 1 \end{aligned}$$

   $$\begin{aligned} \text{Result} &= -\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} - 1 \\ &= -\sqrt{3} - 1 \end{aligned}$$

3. Given $\sin \theta = \frac{3}{5}$ in quadrant II.

   Use the Pythagorean identity to find $\cos \theta$.

   In quadrant II, $\cos$ is negative.

   $$\sin^2 \theta + \cos^2 \theta = 1$$

   $$\left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1$$

   $$\cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}$$

   The value is negative in quadrant II:

   $$\cos\theta=-\frac45$$

   $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{-4/5} = -\frac{3}{4}$$

4. Simplify the angles.

   $$\begin{aligned} &840^\circ - 2(360^\circ) \\ &= 120^\circ \end{aligned}$$

   For $-330^\circ$, add $360^\circ$ to get $30^\circ$.

   $$\begin{aligned} \sin 840^\circ &= \sin 120^\circ \\ &= \sin(180^\circ - 60^\circ) \\ &= \sin 60^\circ \\ &= \frac{\sqrt{3}}{2} \end{aligned}$$

   $$\begin{aligned} \cos(-330^\circ) &= \cos 330^\circ \\ &= \cos(360^\circ - 30^\circ) \\ &= \cos 30^\circ \\ &= \frac{\sqrt{3}}{2} \end{aligned}$$

   $$\begin{aligned} \sin 840^\circ \cdot \cos(-330^\circ) &= \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} \\ &= \frac{3}{4} \end{aligned}$$

5. Angle $1050^\circ$ needs to be simplified first.

   $$\begin{aligned} &1050^\circ - 2(360^\circ) \\ &= 1050^\circ - 720^\circ \\ &= 330^\circ \end{aligned}$$

   Angle $330^\circ$ is in quadrant IV with reference angle $30^\circ$.

   $$x = \cos 330^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}$$

   $$y = \sin 330^\circ = -\sin 30^\circ = -\frac{1}{2}$$

   The blade tip has coordinates $\left(\frac{\sqrt3}{2},-\frac12\right)$.