Exercise 39 Page 963

We want to use a Half-Angle Identity to find the exact value of cos 180^(∘). Let's recall the Half-Angle Identity that involves cosine. cos A/2 = ± sqrt(1+cos A/2) Now, we can use this formula to find the value of cos 180^(∘). We will start by rewriting 180^(∘) as a quotient.

cos 180^(∘)

NumberToFrac

a = 2* a/2

cos (360^(∘)/2)

Substitute

cos A/2= ± sqrt(1+cos A/2)

± sqrt(1+cos 360^(∘)/2)

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 360^(∘) =1. Therefore, we can substitute this value into our expression.

± sqrt(1+cos 360^(∘)/2)

Substitute

cos 360^(∘)= 1}

± sqrt(1+ 1/2)

▼

Simplify

AddTerms

Add terms

± sqrt(2/2)

CalcQuot

Calculate quotient

± sqrt(1)

CalcRoot

Calculate root

± 1

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 180^(∘) is located between quadrants II and III, we know that cos 180^(∘) is negative. cos 180^(∘)= -1