# Nakafa Learning Content > For AI agents: use [llms.txt](https://nakafa.com/llms.txt) for the site index. Markdown versions are available by appending `.md` to content URLs or sending `Accept: text/markdown`. URL: https://nakafa.com/en/subjects/mathematics/function-modeling/trigonometric-function-arbitrary-angle Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/material/lesson/mathematics/function-modeling/trigonometric-function-arbitrary-angle/en.mdx Learn trigonometric functions for any angle using unit circle method. Learn quadrant signs, reference angles, and solve windmill rotation problems. --- ## 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. Visible text: Have you ever observed a clock? When the minute hand moves from to , it forms a angle. Even in one complete rotation, the hand forms a 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 $$O(0,0)$$. Visible text: To understand trigonometric functions of arbitrary angles, we use the unit circle. A circle with a radius of exactly centered at point . Component: UnitCircle Props: - title: Unit Circle Exploration - description: Move the slider to see how coordinates change as the angle rotates. - angle: 30 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 $$(x, y)$$ Visible text: - Angle is always measured from the positive -axis - Positive direction is counterclockwise - Every point on the circle has coordinates **Important definitions:** Component: MathContainer Children: ```math \sin \theta = y \text{ (vertical coordinate)} ``` ```math \cos \theta = x \text{ (horizontal coordinate)} ``` ```math \tan \theta = \frac{y}{x} = \frac{\sin \theta}{\cos \theta}, \quad x \neq 0 ``` ## 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. Visible text: Notice that as the point moves around the circle, the and coordinates can be positive or negative. This is what causes the signs of trigonometric functions to change. Component: Triangle Props: - title: Sign Change Visualization - description: Observe how $$\sin$$, $$\cos$$, and $$\tan$$ values change as the angle passes through each quadrant. Visible text: Observe how , , and values change as the angle passes through each quadrant. - angle: 0 - size: 2 - labels: { opposite: "Opposite Side", adjacent: "Adjacent Side", hypotenuse: "Hypotenuse", } **Signs in each quadrant:** | Quadrant | Angle Range | $$x$$ | $$y$$ | $$\sin$$ | $$\cos$$ | $$\tan$$ | | -------- | ------------------------------------------ | --- | --- | --- | --- | --- | | $$\text{I}$$ | $$0^\circ < \theta < 90^\circ$$ | $$+$$ | $$+$$ | $$+$$ | $$+$$ | $$+$$ | | $$\text{II}$$ | $$90^\circ < \theta < 180^\circ$$ | $$-$$ | $$+$$ | $$+$$ | $$-$$ | $$-$$ | | $$\text{III}$$ | $$180^\circ < \theta < 270^\circ$$ | $$-$$ | $$-$$ | $$-$$ | $$-$$ | $$+$$ | | $$\text{IV}$$ | $$270^\circ < \theta < 360^\circ$$ | $$+$$ | $$-$$ | $$-$$ | $$+$$ | $$-$$ | Visible text: | Quadrant | Angle Range | | | | | | | -------- | ------------------------------------------ | --- | --- | --- | --- | --- | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | To avoid confusion, we can remember this with **"All Students Take Calculus"**. In quadrant I **A**ll are positive, in quadrant II only $$\sin$$ is positive, in quadrant III only $$\tan$$ is positive, in quadrant IV only $$\cos$$ is positive. Visible text: To avoid confusion, we can remember this with **"All Students Take Calculus"**. In quadrant I **A**ll are positive, in quadrant II only is positive, in quadrant III only is positive, in quadrant IV only 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. Visible text: A reference angle is an acute angle ( to ) formed between the terminal side of an angle and the nearest -axis. This concept allows us to use trigonometric values of acute angles that we've already memorized. Component: UnitCircle Props: - title: Understanding Reference Angle - description: Notice the acute angle formed with the $$x$$-axis as the angle changes. Visible text: Notice the acute angle formed with the -axis as the angle changes. - angle: 135 **How to determine reference angle ($$\alpha$$):** Visible text: **How to determine reference angle ():** - Quadrant I: $$\alpha = \theta$$ - Quadrant II: $$\alpha = 180^\circ - \theta$$ - Quadrant III: $$\alpha = \theta - 180^\circ$$ - Quadrant IV: $$\alpha = 360^\circ - \theta$$ Visible text: - Quadrant I: - Quadrant II: - Quadrant III: - Quadrant IV: ## 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 Visible text: 1. **Simplify the angle** (if greater than 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 $$\tan 120^\circ$$ Visible text: **Problem:** Determine , , and **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: $$\sin(+), \cos(-), \tan(-)$$ Visible text: - Angle lies in quadrant II (since ) - Reference angle: - In quadrant II: Using special angle values for $$60^\circ$$: Visible text: Using special angle values for : Component: MathContainer Children: ```math \sin 120^\circ = +\sin 60^\circ = \frac{\sqrt{3}}{2} ``` ```math \cos 120^\circ = -\cos 60^\circ = -\frac{1}{2} ``` ```math \tan 120^\circ = -\tan 60^\circ = -\sqrt{3} ``` ### Angle in Quadrant Three **Problem:** Determine trigonometric values for angle $$240^\circ$$ Visible text: **Problem:** Determine trigonometric values for angle **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: $$\sin(-), \cos(-), \tan(+)$$ Visible text: - Angle lies in quadrant III (since ) - Reference angle: - In quadrant III: Using special angle values for $$60^\circ$$: Visible text: Using special angle values for : Component: MathContainer Children: ```math \sin 240^\circ = -\sin 60^\circ = -\frac{\sqrt{3}}{2} ``` ```math \cos 240^\circ = -\cos 60^\circ = -\frac{1}{2} ``` ```math \tan 240^\circ = +\tan 60^\circ = \sqrt{3} ``` ### Angle in Quadrant Four **Problem:** Determine trigonometric values for angle $$300^\circ$$ Visible text: **Problem:** Determine trigonometric values for angle **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: $$\sin(-), \cos(+), \tan(-)$$ Visible text: - Angle lies in quadrant IV (since ) - Reference angle: - In quadrant IV: Using special angle values for $$60^\circ$$: Visible text: Using special angle values for : Component: MathContainer Children: ```math \sin 300^\circ = -\sin 60^\circ = -\frac{\sqrt{3}}{2} ``` ```math \cos 300^\circ = +\cos 60^\circ = \frac{1}{2} ``` ```math \tan 300^\circ = -\tan 60^\circ = -\sqrt{3} ``` ## 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) Visible text: - (odd function) - (even function) - (odd function) **Example:** $$\sin(-30^\circ) = -\sin 30^\circ = -\frac{1}{2}$$ Visible text: **Example:** ### 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 $$360^\circ$$. Visible text: Use the periodicity property. Subtract or add multiples of until the angle is in the range of to . **Example:** - $$750^\circ = 750^\circ - 2(360^\circ) = 750^\circ - 720^\circ = 30^\circ$$ - Therefore $$\sin 750^\circ = \sin 30^\circ = \frac{1}{2}$$ Visible text: - - Therefore ## 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. Visible text: 1. Determine the values of , , and . 2. Calculate . 3. If and is in quadrant II, determine and . 4. Simplify . 5. A windmill rotates from its initial position. If the initial position of the blade is on the positive -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 $$360^\circ - 315^\circ = 45^\circ$$. ```math \sin 315^\circ = -\sin 45^\circ = -\frac{\sqrt{2}}{2} ``` ```math \cos 315^\circ = +\cos 45^\circ = \frac{\sqrt{2}}{2} ``` ```math \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. ```math \sin(-60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2} ``` ```math \cos 210^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2} ``` ```math \tan(-135^\circ) = -\tan 135^\circ = -(-\tan 45^\circ) = 1 ``` ```math \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. ```math \sin^2 \theta + \cos^2 \theta = 1 ``` ```math \left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1 ``` ```math \cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25} ``` ```math \cos \theta = -\frac{4}{5} \text{ (negative in quadrant II)} ``` ```math \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{-4/5} = -\frac{3}{4} ``` 4. First simplify the angles. ```math 840^\circ = 840^\circ - 2(360^\circ) = 120^\circ ``` For $$-330^\circ$$, add $$360^\circ$$ to get $$30^\circ$$. ```math \sin 840^\circ = \sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} ``` ```math \cos(-330^\circ) = \cos 330^\circ = \cos(360^\circ - 30^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2} ``` ```math \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. ```math 1050^\circ = 1050^\circ - 2(360^\circ) = 1050^\circ - 720^\circ = 330^\circ ``` Angle $$330^\circ$$ is in quadrant IV with reference angle $$30^\circ$$. ```math x = \cos 330^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2} ``` ```math y = \sin 330^\circ = -\sin 30^\circ = -\frac{1}{2} ``` ```math \text{Blade tip coordinates: } \left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right) ``` Visible text: 1. For angle , we need to determine its quadrant first. Since , the angle is in quadrant IV. The reference angle is . 2. Let's calculate each term separately. For , use the odd function property. For , the angle is in quadrant III with reference . For , first convert to positive angle. 3. Given in quadrant II. Use the Pythagorean identity to find . Remember that in quadrant II, is negative. 4. First simplify the angles. For , add to get . 5. Angle needs to be simplified first. Angle is in quadrant IV with reference angle .