Exercise 35 Page 963

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

cos 60^(∘)

SplitIntoFactors

Split into factors

cos ( 2* 30^(∘))

Substitute

cos 2 θ= 2cos^2 θ- 1

2cos^2 30^(∘) -1

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 30^(∘) = sqrt(3)2. Therefore, we can substitute this value into our expression.

2cos^2 30^(∘) -1

Substitute

cos 30^(∘)= sqrt(3)/2

2 (sqrt(3)/2)^2 - 1

▼

Simplify

PowQuot

(a/b)^m=a^m/b^m

2 ( sqrt(3)^2/2^2 )- 1

PowSqrt

( sqrt(a) )^2 = a

2 ( 3/2^2) - 1

CalcPow

Calculate power

2 ( 3/4 )- 1

MoveLeftFacToNum

a*b/c= a* b/c

6/4 - 1

ReduceFrac

a/b=.a /2./.b /2.

3/2 - 1

NumberToFrac

a = * a/

3/2 - 2/2

SubFrac

Subtract fractions

1/2