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Finding Missing Angles Using the Unit Circle

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Finding Missing Angles Using the Unit Circle

Finding Missing Angles Using the Unit Circle

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Question 1 of 3

1. Question

Given that

0 \leq θ \leq 2 π

$0≤theta≤2pi$ , solve for

θ

$theta$

{tan}^{2} θ = 3

$tan^2theta=3$

1. $\frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6}$ $pi/6,(5pi)/6,(7pi)/6,(11pi)/6$
2. $\frac{π}{3}, \frac{4 π}{3}$ $pi/3,(4pi)/3$
3. $\frac{π}{4}, \frac{7 π}{6}$ $pi/4,(7pi)/6$
4. $\frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}$ $pi/3,(2pi)/3,(4pi)/3,(5pi)/3$

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Trigonometric Functions

sin θ = \frac{opposite}{hypotenuse}

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$

cos θ = \frac{adjacent}{hypotenuse}

$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$

tan θ = \frac{opposite}{adjacent}

$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

First, simplify the given equation

${tan}^{2} θ$ $tan^2theta$	$=$ $=$	$3$ $3$
$\sqrt{{tan}^{2} θ}$ $sqrt(tan^2theta)$	$=$ $=$	$\sqrt{3}$ $sqrt3$	Get the square root of both sides
$tan θ$ $tan theta$	$=$ $=$	$\pm \sqrt{3}$ $+-sqrt3$

Remember that

tan θ = \frac{sin θ}{cos θ}

$tan theta=(sin theta)/(cos theta)$ . Knowing that

tan θ

$tan theta$ can be either positive or negative, this means that

θ

$theta$ lies on all

4

$4$ quadrants

Next, find the reference angle to work with by using special triangles

$tan θ$ $tan theta$	$=$ $=$	$\pm \sqrt{3}$ $+-sqrt3$	$=$ $=$	$\frac{\text{opposite}}{\text{adjacent}}$

$tan θ$ $tan theta$	$=$ $=$	$\pm \sqrt{3}$ $+-sqrt3$	$=$ $=$	$\frac{\sqrt{3}}{1}$ $(sqrt3)/1$	Using $\frac{π}{3}$ $pi/3$ as $θ$ $theta$

$tan θ$ $tan theta$	$=$ $=$	$\pm \sqrt{3}$ $+-sqrt3$	$=$ $=$	$\sqrt{3}$ $sqrt3$

Hence, the reference angle and the value of

θ

$theta$ on the First Quadrant is $\frac{π}{3}$ $pi/3$

Finally, use the reference angle to find the value of

θ

$theta$ on the other

3

$3$ quadrants

Remember that the reference angle is always relative to the

x

$x$ -axis

Second Quadrant:

$θ$ $theta$	$=$ $=$	$π - \frac{π}{3}$ $pi-pi/3$

	$=$ $=$	$\frac{3 π}{3} - \frac{π}{3}$ $(3pi)/3-pi/3$

	$=$ $=$	$\frac{2 π}{3}$ $(2pi)/3$

Third Quadrant:

$θ$ $theta$	$=$ $=$	$π + \frac{π}{3}$ $pi+pi/3$

	$=$ $=$	$\frac{3 π}{3} + \frac{π}{3}$ $(3pi)/3+pi/3$

	$=$ $=$	$\frac{4 π}{3}$ $(4pi)/3$

Fourth Quadrant:

$θ$ $theta$	$=$ $=$	$2 π - \frac{π}{3}$ $2pi-pi/3$

	$=$ $=$	$\frac{6 π}{3} - \frac{π}{3}$ $(6pi)/3-pi/3$

	$=$ $=$	$\frac{5 π}{3}$ $(5pi)/3$

θ = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}

$theta=pi/3,(2pi)/3,(4pi)/3,(5pi)/3$

Question 2 of 3

2. Question
Given that $0 \leq θ \leq 2 π$ $0≤theta≤2pi$ , solve for $θ$ $theta$

${sin}^{2} θ = 1$ $sin^2theta=1$
- 1. $\frac{π}{2}, π, \frac{3 π}{2}, 2 π$ $pi/2,pi,(3pi)/2,2pi$
- 2. $\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ $pi/4,(3pi)/4,(5pi)/4,(7pi)/4$
- 3. $\frac{π}{4}, \frac{3 π}{4}$ $pi/4,(3pi)/4$
- 4. $\frac{π}{2}, \frac{3 π}{2}$ $pi/2,(3pi)/2$
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Trigonometric Functions

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$

$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$

$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

First, simplify the given equation

${sin}^{2} θ$ $sin^2theta$ $=$ $=$ $1$ $1$

$\sqrt{{sin}^{2} θ}$ $sqrt(sin^2theta)$ $=$ $=$ $\sqrt{1}$ $sqrt1$ Get the square root of both sides

$sin θ$ $sin theta$ $=$ $=$ $\pm 1$ $+-1$

In the unit circle. mark out the coordinates on the horizontal and vertical axis

Keep in mind that the radius of a unit circle is $1$ $1$

Now, recall that in a coordinate, $(x, y) = (cos θ, sin θ)$ $(x,y)=(cos theta, sin theta)$

$sin θ = \pm 1$ $sin theta=+-1$

The $y$ $y$ value in these coordinates match the value of $sin θ$ $sin theta$ . Therefore, $θ = \frac{π}{2}, \frac{3 π}{2}$ $theta=pi/2,(3pi)/2$

$θ = \frac{π}{2}, \frac{3 π}{2}$ $theta=pi/2,(3pi)/2$
Question 3 of 3

3. Question
Given that $0 \leq θ \leq 2 π$ $0≤theta≤2pi$ , solve for $θ$ $theta$

$cos θ = - 1$ $costheta=-1$
- 1. $θ = 2 π$ $theta=2pi$
- 2. $θ = \frac{3 π}{2}$ $theta=(3pi)/2$
- 3. $θ = \frac{π}{2}$ $theta=pi/2$
- 4. $θ = π$ $theta=pi$
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Trigonometric Functions

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$

$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$

$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

In the unit circle. mark out the coordinates on the horizontal and vertical axis

Keep in mind that the radius of a unit circle is $1$ $1$

Now, recall that in a coordinate, $(x, y) = (cos θ, sin θ)$ $(x,y)=(cos theta, sin theta)$

$cos θ = - 1$ $cos theta=-1$

The $x$ $x$ value in $π$ $pi$ the value of $cos θ$ $cos theta$ . Therefore, $θ = π$ $theta=pi$

$θ = π$ $theta=pi$

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

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