Unit Circle: Exact Values 1

Question 1 of 4

Find the exact value of:

$sin330°$

1. $-(sqrt3)/2$
2. $1/sqrt3$
3. $-1/2$
4. $2$

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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: All

Quadrant II: Sine only

Quadrant III: Tangent only

Quadrant IV: Cosine only

First, 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

Find the exact value of:

$tan120°$

1. $1/2$
2. $- 1/sqrt3$
3. $2/sqrt3$
4. $-sqrt3$

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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: All

Quadrant II: Sine only

Quadrant III: Tangent only

Quadrant IV: Cosine only

First, 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

Find the exact value of:

$sec315°$

1. $1/2$
2. $sqrt2$
3. $-sqrt2$
4. $1/sqrt2$

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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: All

Quadrant II: Sine only

Quadrant III: Tangent only

Quadrant IV: Cosine only

First, 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

Find the exact value of:

$cot150°$

1. $-sqrt3$
2. $2/sqrt3$
3. $1/sqrt3$
4. $- 1/sqrt3$

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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: All

Quadrant II: Sine only

Quadrant III: Tangent only

Quadrant IV: Cosine only

First, 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

$cot theta = 1/(tan theta)$

$cot150°$	$=$	$-cot30°$

	$=$	$- 1/(tan30°)$

	$=$	$-\frac{1}{\frac{\color{#9a00c7}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}}$

	$=$	$-\frac{1}{\frac{\color{#9a00c7}{1}}{\color{#00880A}{\sqrt3}}}$

	$=$	$-sqrt3$

$-sqrt3$

Unit Circle: Exact Values 1

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Quiz summary

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1. Question

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Excellent!

Trigonometric Functions

Positive Values in the Unit Circle

2. Question

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Trigonometric Functions

Positive Values in the Unit Circle

3. Question

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Trigonometric Functions

Positive Values in the Unit Circle

4. Question

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Trigonometric Functions

Positive Values in the Unit Circle

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