Trigonometric Function of Arbitrary Angle

Understanding Angles Greater than a Right Angle

Have you ever observed a clock? When the minute hand moves from $12$ to $6$, it forms a $180^\circ$ angle. Even in one complete rotation, the hand forms a $360^\circ$ angle.

In mathematics, we need to understand trigonometric values for angles like these. Not just limited to acute angles in right triangles.

Unit Circle

To understand trigonometric functions of arbitrary angles, we use the unit circle. A circle with a radius of exactly $1 \text{ unit}$ centered at point $O(0,0)$.

Let's understand in detail:

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

Important definitions:

Why Do Signs Change in Each Quadrant?

Notice that as the point moves around the circle, the $x$ and $y$ coordinates can be positive or negative. This is what causes the signs of trigonometric functions to change.

Signs in each quadrant:

To avoid confusion, we can remember this with "All Students Take Calculus". In quadrant I All are positive, in quadrant II only $\sin$ is positive, in quadrant III only $\tan$ is positive, in quadrant IV only $\cos$ is positive.

Reference Angle

A reference angle is an acute angle ($0^\circ$ to $90^\circ$) formed between the terminal side of an angle and the nearest $x$-axis. This concept allows us to use trigonometric values of acute angles that we've already memorized.

How to determine reference angle ($\alpha$):

• 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

Negative Angles

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:

• $750^\circ = 750^\circ - 2(360^\circ) = 750^\circ - 720^\circ = 30^\circ$
• Therefore $\sin 750^\circ = \sin 30^\circ = \frac{1}{2}$

Exercises

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. Let's 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.

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

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

   $$\tan(-135^\circ) = -\tan 135^\circ = -(-\tan 45^\circ) = 1$$

   $$\text{Result} = -\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} + 1 = -\sqrt{3} + 1$$

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

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

   Remember that in quadrant II, $\cos$ is negative.

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

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

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

   $$\cos \theta = -\frac{4}{5} \text{ (negative in quadrant II)}$$

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

4. First simplify the angles.

   $$840^\circ = 840^\circ - 2(360^\circ) = 120^\circ$$

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

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

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

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

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

   $$1050^\circ = 1050^\circ - 2(360^\circ) = 1050^\circ - 720^\circ = 330^\circ$$

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

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

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

   $$\text{Blade tip coordinates: } \left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$$