Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
6. Angle Identities
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Exercise 18 Page 949

Use the Negative Angle Identity tan (- θ)=-tan θ and the Cofunction Identity tan ( π2-θ)=cot θ.

π/4, 3π/4, 5π/4, and 7π/4

Practice makes perfect
Let's start by recalling the Negative Angle Identity and the Cofunction Identity that we will use. Cofunction Identity tan (π/2-θ)=cot θ [0.8em] Negative Angle Identity tan (- θ)=-tan θ To solve the given equation, we will first rewrite it using these identities.
tan (π/2-θ ) + tan(-θ)=0
cot θ + tan(-θ)=0
cot θ + ( -tan θ)=0
cot θ - tan θ=0
Now, we will rewrite cot θ using the Cotangent Identity.
cot θ - tan θ=0
1/tan θ - tan θ=0
Solve for tan θ
tan θ * 1/tan θ - tan θ * tan θ =0 * tan θ
1 - tan θ * tan θ =0 * tan θ
1 - tan^2 θ =0 * tan θ
1 - tan^2 θ =0
-tan^2 θ =-1
tan^2 θ =1
sqrt(tan^2 θ) =sqrt(1)
tan θ =± sqrt(1)
tan θ = ± 1
lc tan θ = 1 & (I) tan θ = -1 & (II)

We will solve the obtained equations one at a time.

tan θ = 1

Next, we will isolate θ. tan θ =1 ⇔ θ =tan ^(- 1) 1 Let's use a calculator to find one value for θ.
θ =tan ^(- 1)1
θ =π/4
Finally, we need to check if there are any other possible angles that satisfy this equation within the given range. Given Range: 0 ≤ θ < 2π Recall that the tangent of an angle in standard position intersects the unit circle at the point (x,y). Let's plot all the points on the unit circle where the ratio yx equals 1.

As we can see above, the angles whose tangent is 1 are π4 and 5π4.

tan θ = -1

Next, we will isolate θ. tan θ =-1 ⇔ θ =tan ^(- 1) (-1) Let's use a calculator to find one value for θ.
θ =tan ^(- 1)(-1)
θ =3π/4
Finally, we need to check if there are any other possible angles that satisfy this equation within the given range. Given Range: 0 ≤ θ < 2π Recall that the tangent of an angle in standard position intersects the unit circle at the point (x,y). Let's plot all the points on the unit circle where the ratio yx equals -1.

As we can see above, the angles whose tangent is -1 are 3π4 and 7π4. Therefore, the solutions for the equation in the given range are π4, 5π4, 3π4, and 7π4.