Exercise 37 Page 963

We want to use a Half-Angle Identity to find the exact value of tan 30 ^(∘). Let's recall the Half-Angle Identity that involves cosine. tan A/2 = ± sqrt(1-cos A/1+ cos A) Now, we can use this formula to find the value of tan 30^(∘). We will start by rewriting 30^(∘) as a quotient.

tan 30^(∘)

NumberToFrac

a = 2* a/2

tan (60^(∘)/2)

Substitute

tan A/2= ± sqrt(1-cos A/1+cos A)

± sqrt(1-cos 60^(∘)/1+cos 60^(∘))

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

We can see in the table that cos 60^(∘) = 12. Therefore, we can substitute this value into our expression.

± sqrt(1-cos 60^(∘)/1+cos 60^(∘))

Substitute

cos 60^(∘)= 1/2

± sqrt(1- 12/1+ 12)

▼

Simplify

OneToFrac

Rewrite 1 as 2/2

± sqrt(22 - 12/22 + 12)

AddSubFrac

Add and subtract fractions

± sqrt(12/32)

DivFracByFracD

.a /b./.c /d.=a/b*d/c

± sqrt(1/3)

ExpandFrac

a/b=a * 3/b * 3

± sqrt(3/9)

SqrtQuot

sqrt(a/b)=sqrt(a)/sqrt(b)

± sqrt(3)/sqrt(9)

CalcRoot

Calculate root

± sqrt(3)/3

Finally, we will determine the sign. To do so, let's recall the signs of sine, cosine, and tangent in the four quadrants of the coordinate plane.

Since 30^(∘) is located in Quadrant I, we know that tan 30^(∘) is positive. tan 30^(∘)= sqrt(3)/3