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Pearson Algebra 2 Common Core, 2011 View details

Chapter Test

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Exercise 36 Page 963

Use the Double Angle Identity tan 2θ = 2 tan θ 1- tan^2 θ.

sqrt(3)

Practice makes perfect

We want to use a Double Angle Identity to find the exact value of tan 60^(∘). Let's recall the Double Angle Identity that involves sine. tan 2θ = 2 tan θ/1- tan^2 θ Now, we can use this formula to find the value of tan 60^(∘). We will start by rewriting 60^(∘) as a product.

tan 60^(∘)

SplitIntoFactors

Split into factors

tan ( 2* 30^(∘))

Substitute

tan 2 θ= 2tan θ/1-tan^2 θ

2tan 30^(∘)/1-tan^2 30^(∘)

Next we will recall the value of sine, cosine, and tangent for some special angles.

Trigonometric Values for Special Angles
Sine	Cosine	Tangent
sin 0^(∘)=0	cos 0^(∘)=1	tan 0^(∘)=0
sin 30^(∘)=1/2	cos 30^(∘)=sqrt(3)/2	tan 30^(∘)=sqrt(3)/3
sin 60^(∘)=sqrt(3)/2	cos 60^(∘)=1/2	tan 60^(∘)=sqrt(3)
sin 90^(∘) = 1	cos 90^(∘) = 0	-
sin 120^(∘)= sqrt(3)/2	cos 120^(∘)= - 1/2	tan 120^(∘)= - sqrt(3)
sin 150^(∘)= 1/2	cos 150^(∘)= - sqrt(3)/2	tan 150^(∘)= - sqrt(3)/3
sin 180^(∘)= 0	cos 180^(∘)= - 1	tan 180^(∘)= 0
sin 210^(∘)= - 1/2	cos 210^(∘)= - sqrt(3)/2	tan 210^(∘)= sqrt(3)/3
sin 240^(∘)= - sqrt(3)/2	cos 240^(∘)= - 1/2	tan 240^(∘)= sqrt(3)
sin 270^(∘)= - 1	cos 270^(∘)= 0	-
sin 300^(∘)= - sqrt(3)/2	cos 300^(∘)= 1/2	tan 300^(∘)= - sqrt(3)
sin 330^(∘) = - 1/2	cos 330^(∘) = sqrt(3)/2	tan 330^(∘) = - sqrt(3)/3
sin 360^(∘)= 0	cos 360^(∘)= 1	tan 360^(∘)= 0