Trig Exact Values 1
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Question 1 of 4
1. Question
Find the exact value of:`sin330°`Hint
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Trigonometric Functions
$$\sin\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#007DDC}{\text{hypotenuse}}}$$$$\cos\theta=\frac{\color{#00880A}{\text{adjacent}}}{\color{#007DDC}{\text{hypotenuse}}}$$$$\tan\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$$
Positive Values in the Unit Circle
Quadrant I: AllQuadrant II: Sine onlyQuadrant III: Tangent onlyQuadrant IV: Cosine onlyFirst, draw in the ray of `330°`.
Next, get the acute angle (`theta’`) for `330°`.The acute angle is an angle less than `90°` and is relative to the horizontal axis.
Based on this diagram, simply subtract `330°` from `360°` to get the acute angle.`theta’` `=` `360°-330°` `=` `30°` Now, identify if sine is positive or negative in the quadrant where the ray lies, Quadrant IV.
The sine value is negative in Quadrant IV.Finally, we can get the exact value of `sin330°` by solving for `-sin30°` using Exact Triangle Ratios.
`sin330°` `=` `-sin30°` `=` $$-\frac{\color{#9a00c7}{\text{opposite}}}{\color{#007DDC}{\text{hypotenuse}}}$$ `=` $$-\frac{\color{#9a00c7}{1}}{\color{#007DDC}{2}}$$ `-1/2` -
Question 2 of 4
2. Question
Find the exact value of:`tan120°`Hint
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Trigonometric Functions
$$\sin\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#007DDC}{\text{hypotenuse}}}$$$$\cos\theta=\frac{\color{#00880A}{\text{adjacent}}}{\color{#007DDC}{\text{hypotenuse}}}$$$$\tan\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$$Positive Values in the Unit Circle
Quadrant I: AllQuadrant II: Sine onlyQuadrant III: Tangent onlyQuadrant IV: Cosine onlyFirst, draw in the ray of `120°`.
Next, get the acute angle (`theta’`) for `120°`.The acute angle is an angle less than `90°` and is relative to the horizontal axis.
Based on this diagram, simply subtract `120°` from `180°` to get the acute angle.`theta’` `=` `180°-120°` `=` `60°` Now, identify if tangent is positive or negative in the quadrant where the ray lies, Quadrant II.
The tangent value is negative in Quadrant II.Finally, we can get the exact value of `tan120°` by solving for `-tan60°` using Exact Triangle Ratios.
`tan120°` `=` `-tan60°` `=` $$-\frac{\color{#9a00c7}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$$ `=` $$-\frac{\color{#9a00c7}{\sqrt3}}{\color{#00880A}{1}}$$ `=` `-sqrt3` `-sqrt3` -
Question 3 of 4
3. Question
Find the exact value of:`sec315°`Hint
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Trigonometric Functions
$$\sin\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#007DDC}{\text{hypotenuse}}}$$$$\cos\theta=\frac{\color{#00880A}{\text{adjacent}}}{\color{#007DDC}{\text{hypotenuse}}}$$$$\tan\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$$Positive Values in the Unit Circle
Quadrant I: AllQuadrant II: Sine onlyQuadrant III: Tangent onlyQuadrant IV: Cosine onlyFirst, draw in the ray of `315°`.
Next, get the acute angle (`theta’`) for `315°`.The acute angle is an angle less than `90°` and is relative to the horizontal axis.
Based on this diagram, simply subtract `315°` from `360°` to get the acute angle.`theta’` `=` `360°-315°` `=` `45°` Finally, we can get the exact value of `sec315°` by solving for `sec45°` using Exact Triangle Ratios.Note that since `sec theta=1/(cos theta)` and cosine is positive in Quadrant IV, secant is also positive.
`sec315°` `=` `sec45°` `=` `1/(cos45°)` `=` $$\frac{1}{\frac{\color{#9a00c7}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}}$$ `=` $$\frac{1}{\frac{\color{#9a00c7}{1}}{\color{#00880A}{\sqrt2}}}$$ `=` `sqrt2` `sqrt2` -
Question 4 of 4
4. Question
Find the exact value of:`cot150°`Hint
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Great Work!
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Trigonometric Functions
$$\sin\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#007DDC}{\text{hypotenuse}}}$$$$\cos\theta=\frac{\color{#00880A}{\text{adjacent}}}{\color{#007DDC}{\text{hypotenuse}}}$$$$\tan\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$$Positive Values in the Unit Circle
Quadrant I: AllQuadrant II: Sine onlyQuadrant III: Tangent onlyQuadrant IV: Cosine onlyFirst, draw in the ray of `150°`.
Next, get the acute angle (`theta’`) for `150°`.The acute angle is an angle less than `90°` and is relative to the horizontal axis.
Based on this diagram, simply subtract `150°` from `180°` to get the acute angle.`theta’` `=` `180°-150°` `=` `30°` Now, identify if cotangent is positive or negative in the quadrant where the ray lies, Quadrant II.
The cotangent value is negative in Quadrant II.Finally, we can get the exact value of `cot150°` by solving for `-cot30°` using Exact Triangle Ratios.Note that `cot theta = 1/(tan theta)`
`cot150°` `=` `-cot30°` `=` `- 1/(tan30°)` `=` $$-\frac{1}{\frac{\color{#9a00c7}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}}$$ `=` $$-\frac{1}{\frac{\color{#9a00c7}{1}}{\color{#00880A}{\sqrt3}}}$$ `=` `-sqrt3` `-sqrt3`
Quizzes
- Converting Angle Measures 1
- Converting Angle Measures 2
- Converting Angle Measures 3
- Finding the Central Angle in a Circle
- Finding Areas in a Circle
- Values on the Unit Circle
- Finding Missing Angles Using the Unit Circle
- Trigonometric Ratios in the Unit Circle
- Trig Exact Values 1
- Trig Exact Values 2
- Trig Equations
- Derivative of a Trigonometric Function 1
- Derivative of a Trigonometric Function 2
- Derivative of a Trigonometric Function 3
- Applications of Differentiation
- Integral of a Trigonometric Function 1
- Integral of a Trigonometric Function 2
- Applications of Integration
- Graphing Trigonometric Functions 1
- Graphing Trigonometric Functions 2
- Graphing Trigonometric Functions 3
- Graphing Trigonometric Functions 4