Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
2. Solving Trigonometric Equations Using Inverses
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Exercise 10 Page 916

The tangent of an angle is defined as the ratio of the sine to the cosine of the angle.

120^(∘) + n* 360^(∘) and 300^(∘) + n* 360^(∘), for any integer n

Practice makes perfect
We want to find all the angles whose tangent is -sqrt(3). Since we know the value of tangent but not the value of the angle, the process of finding the missing value uses an inverse trigonometric function. tan^(- 1)(- sqrt(3)) First, we need to use the definition of tangent to find the appropriate reference angle. The tangent of an angle is defined as the ratio of the sine to the cosine of the angle. tan θ =sin θ/cos θ Note that we will actually find the opposite of the desired angle because all of the reference angles have positive values for sine and cosine while the given tangent value is negative.
sin θ cos θ sin θ/cos θ tan θ
θ=30^(∘) 1/2 sqrt(3)/2 12/sqrt(3)2 sqrt(3)/3
θ=45^(∘) sqrt(2)/2 sqrt(2)/2 sqrt(2)2/sqrt(2)2 1
θ=60^(∘) sqrt(3)/2 1/2 sqrt(3)2/12 sqrt(3)
θ=90^(∘) 1 0 1/0 Undefined

The tangent of an angle that measures 60^(∘) is sqrt(3). Let's now construct a 30^(∘)-60^(∘)-90^(∘) triangle. Since we will also need to use the unit circle, our right triangle will have a hypotenuse that measures 1.

30-60-90 triangle

In this type of special triangle, the length of the shorter leg is half the length of the hypotenuse, and the length of the longer leg is sqrt(3) times the length of the shorter leg.

30-60-90 triangle

Next, let's locate this triangle in the coordinate plane. Since the given value is negative, our solutions will be located in the quadrants that have negative tangent values — Quadrant II and Quadrant IV.

tangent

We can draw a 30^(∘)-60^(∘)-90^(∘) triangle on a unit circle with the measurements found above in these quadrants. We will place the 60^(∘) angle at the origin.

unit circles and triangles

Finally, we will use these triangles to locate two angles whose tangent is - sqrt(3). Keep in mind that a full turn measures 360^(∘) and a half turn measures 180^(∘). Therefore, we will subtract 60 from 180 and from 360. 180- 60&= 120^(∘) 360- 60&= 300^(∘) Let's show the obtained angles on the coordinate plane.

We found that tan(- sqrt(3)) can be either 120^(∘) or 300^(∘). Finally, keep in mind that if we add or subtract a multiple of 360^(∘), the terminal side of the angle will be in the same position. Therefore, the tangent of the resulting angles will also be - sqrt(3). 120^(∘)+ n * 360^(∘) and 300^(∘)+ n * 360^(∘), wheren is any integer