Unit Circle: Find Angles from Equations

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Question 1 of 3

1. Question
Given that $0 \geq θ \geq 360$ $0≥theta≥360$ , solve:

$cos θ = - \frac{1}{\sqrt{2}}$ $cos theta=- 1/sqrt2$
- $θ =$ $theta=$ (135, 225) $°,$ $°,$ (225, 135) $°$ $°$
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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

Exact Triangle Ratios

The acute angle $(θ ’)$ $(theta’)$ is an angle less than $90 °$ $90°$ and is relative to the horizontal axis.

First, identify which quadrant $θ$ $theta$ may lie.

Since the given cosine value is negative, $θ$ $theta$ may be on Quadrant II or III.

Next, get the acute angle ( $θ ’$ $theta’$ ) for $θ$ $theta$ .

We can do this by matching the value of $cos θ$ $cos theta$ to an Exact Triangle.

$\cos\theta=\frac{\color{#00880A}{\text{adjacent}}}{\color{#007DDC}{\text{hypotenuse}}}=-\frac{\color{#00880A}{1}}{\color{#007DDC}{\sqrt2}}$

Since $1$ $1$ is the adjacent side and $\sqrt{2}$ $sqrt2$ is the hypotenuse, $θ ’ = 45 °$ $theta’=45°$

Now, find the corresponding angle of the acute angle in both Quadrant II and III.

Quadrant II:

$180 ° - 45 ° =$ $180°-45°=$ $135 °$ $135°$

Quadrant III:

$180 ° + 45 ° =$ $180°+45°=$ $225 °$ $225°$

Therefore, $θ = 135 °, 225 °$ $theta=135°,225°$

$θ = 135 °, 225 °$ $theta=135°,225°$
Question 2 of 3

2. Question
Given that $0 \geq θ \geq 360$ $0≥theta≥360$ , solve:

$sin θ = \frac{\sqrt{3}}{2}$ $sin theta=(sqrt3)/2$
- $θ =$ $theta=$ (60, 120) $°,$ $°,$ (120, 60) $°$ $°$
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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

Exact Triangle Ratios

The acute angle $(θ ’)$ $(theta’)$ is an angle less than $90 °$ $90°$ and is relative to the horizontal axis.

First, identify which quadrant $θ$ $theta$ may lie.

Since the given sine value is positive, $θ$ $theta$ may be on Quadrant I or II.

Next, get the acute angle ( $θ ’$ $theta’$ ) for $θ$ $theta$ .

We can do this by matching the value of $sin θ$ $sin theta$ to an Exact Triangle.

$\sin\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#007DDC}{\text{hypotenuse}}}=\frac{\color{#9a00c7}{\sqrt3}}{\color{#007DDC}{2}}$

Since $\sqrt{3}$ $sqrt3$ is the opposite side and $2$ $2$ is the hypotenuse, $θ ’ = 60 °$ $theta’=60°$

Now, find the corresponding angle of the acute angle in both Quadrant I and II.

Quadrant I:

$60 °$ $60°$

Quadrant II:

$180 ° - 60 ° =$ $180°-60°=$ $120 °$ $120°$

Therefore, $θ = 60 °, 120 °$ $theta=60°,120°$

$θ = 60 °, 120 °$ $theta=60°,120°$
Question 3 of 3

3. Question
Given that $0 \geq x \geq 360$ $0≥x≥360$ , solve:

$tan x = - 1$ $tan x=-1$
- $x =$ $x=$ (135, 315) $°,$ $°,$ (135, 315) $°$ $°$
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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

Exact Triangle Ratios

The acute angle $(θ ’)$ $(theta’)$ is an angle less than $90 °$ $90°$ and is relative to the horizontal axis.

First, identify which quadrant $x$ $x$ may lie.

Since the given tangent value is negative, $x$ $x$ may be on Quadrant II or IV.

Next, get the acute angle for $x$ $x$ .

We can do this by matching the value of $tan x$ $tan x$ to an Exact Triangle.

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

Since $1$ $1$ is the opposite side and $1$ $1$ is the adjacent, the acute angle is $45 °$ $45°$

Now, find the corresponding angle of the acute angle in both Quadrant II and IV.

Quadrant II:

$180 ° - 45 ° =$ $180°-45°=$ $135 °$ $135°$

Quadrant IV:

$360 ° - 45 ° =$ $360°-45°=$ $315 °$ $315°$

Therefore, $x = 135 °, 315 °$ $x=135°,315°$

$x = 135 °, 315 °$ $x=135°,315°$

Unit Circle: Find Angles from Equations

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Quiz summary

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1. Question

Hint

Well Done!

Trigonometric Functions

Positive Values in the Unit Circle

Exact Triangle Ratios

2. Question

Hint

Fantastic!

Trigonometric Functions

Positive Values in the Unit Circle

Exact Triangle Ratios

3. Question

Hint

Great Work!

Trigonometric Functions

Positive Values in the Unit Circle

Exact Triangle Ratios

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