Trig Exact Values 2

Question 1 of 4

Find the exact value of:

$sin(-210)°$

Write fractions in the format “a/b”

(1/2)

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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

$-210°$ .

A negative angle goes anti-clockwise.

Next, get the acute angle (

$theta’$ ) for

$-210°$ .

The acute angle is an angle less than

$90°$ and is relative to the horizontal axis.

Based on this diagram, simply subtract

$180°$ from

$210°$ to get the acute angle.

$theta’$	$=$	$210°-180°$
	$=$	$30°$

Finally, we can get the exact value of

$sin(-210)°$ by solving for

$sin30°$ using Exact Triangle Ratios.

Note that the value is positive because Sine is positive on Quadrant II.

$sin(-210)°$	$=$	$sin30°$

	$=$	$\frac{\color{#9a00c7}{\text{opposite}}}{\color{#007DDC}{\text{hypotenuse}}}$

	$=$	$\frac{\color{#9a00c7}{1}}{\color{#007DDC}{2}}$

$1/2$

Question 2 of 4

Find the exact value of:

$cos495°$

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

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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

$495°$ .

Since the angle is greater than

$360°$ , it makes a full revolution. To draw the ray, we must first subtract

$360°$ .

$495°-360°=135°$

Next, get the acute angle (

$theta’$ ) for

$495°$ .

The acute angle is an angle less than

$90°$ and is relative to the horizontal axis.

Based on this diagram, simply subtract

$135°$ from

$180°$ to get the acute angle.

$theta’$	$=$	$180°-135°$
	$=$	$45°$

Now, identify if cosine is positive or negative in the quadrant where the ray lies, Quadrant II.

The cosine value is negative in Quadrant II.

Finally, we can get the exact value of

$cos495°$ by solving for

$-cos45°$ using Exact Triangle Ratios.

$cos495°$	$=$	$-cos45°$

	$=$	$-\frac{\color{#00880A}{\text{adjacent}}}{\color{#007DDC}{\text{hypotenuse}}}$

	$=$	$-\frac{\color{#00880A}{1}}{\color{#007DDC}{\sqrt2}}$

$- 1/sqrt2$

Question 3 of 4

Find the exact value of:

$tan150°$

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.

$theta’$	$=$	$180°-150°$
	$=$	$30°$

Finally, we can get the exact value of

$tan150°$ by solving for

$-tan30°$ using Exact Triangle Ratios.

Note that the value is negative because tangent is negative in Quadrant II.

$tan150°$	$=$	$-tan30°$

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

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

$- 1/sqrt3$

Question 4 of 4

Find the exact value of:

$cos315°$

1. $sqrt2$
2. $sqrt3/2$
3. $- 1/sqrt2$
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

$cos315°$ by solving for

$cos45°$ using Exact Triangle Ratios.

Note that the value is positive because cosine is positive on Quadrant IV.

$cos315°$	$=$	$cos45°$

	$=$	$\frac{\color{#00880A}{\text{adjacent}}}{\color{#007DDC}{\text{hypotenuse}}}$

	$=$	$\frac{\color{#00880A}{1}}{\color{#007DDC}{\sqrt2}}$

$1/sqrt2$

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Quiz summary

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1. Question

Hint

Good Job!

Trigonometric Functions

Positive Values in the Unit Circle

2. Question

Hint

Excellent!

Trigonometric Functions

Positive Values in the Unit Circle

3. Question

Hint

Nice Job!

Trigonometric Functions

Positive Values in the Unit Circle

4. Question

Hint

Exceptional!

Trigonometric Functions

Positive Values in the Unit Circle

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