Unit Circle: Find Angles
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Question 1 of 3
1. Question
Given that `0≤theta≤2pi`, solve for `theta``tan^2theta=3`Hint
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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`tan^2theta` `=` `3` `sqrt(tan^2theta)` `=` `sqrt3` Get the square root of both sides `tan theta` `=` `+-sqrt3` Remember that `tan theta=(sin theta)/(cos theta)`. Knowing that `tan theta` can be either positive or negative, this means that `theta` lies on all `4` quadrants
Next, find the reference angle to work with by using special triangles
`tan theta` `=` `+-sqrt3` `=` $$\frac{\text{opposite}}{\text{adjacent}}$$ `tan theta` `=` `+-sqrt3` `=` `(sqrt3)/1` Using `pi/3` as `theta` `tan theta` `=` `+-sqrt3` `=` `sqrt3` Hence, the reference angle and the value of `theta` on the First Quadrant is `pi/3`
Finally, use the reference angle to find the value of `theta` on the other `3` quadrantsRemember that the reference angle is always relative to the `x`-axisSecond Quadrant:
`theta` `=` `pi-pi/3` `=` `(3pi)/3-pi/3` `=` `(2pi)/3` Third Quadrant:
`theta` `=` `pi+pi/3` `=` `(3pi)/3+pi/3` `=` `(4pi)/3` Fourth Quadrant:
`theta` `=` `2pi-pi/3` `=` `(6pi)/3-pi/3` `=` `(5pi)/3` `theta=pi/3,(2pi)/3,(4pi)/3,(5pi)/3` -
Question 2 of 3
2. Question
Given that `0≤theta≤2pi`, solve for `theta``sin^2theta=1`Hint
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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^2theta` `=` `1` `sqrt(sin^2theta)` `=` `sqrt1` Get the square root of both sides `sin theta` `=` `+-1` In the unit circle. mark out the coordinates on the horizontal and vertical axisKeep in mind that the radius of a unit circle is `1`
Now, recall that in a coordinate, `(x,y)=(cos theta, sin theta)``sin theta=+-1`
The `y` value in these coordinates match the value of `sin theta`. Therefore, `theta=pi/2,(3pi)/2``theta=pi/2,(3pi)/2` -
Question 3 of 3
3. Question
Given that `0≤theta≤2pi`, solve for `theta``costheta=-1`Hint
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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 axisKeep in mind that the radius of a unit circle is `1`
Now, recall that in a coordinate, `(x,y)=(cos theta, sin theta)``cos theta=-1`
The `x` value in `pi` the value of `cos theta`. Therefore, `theta=pi``theta=pi`