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Section 3.3 Double-Angle and Half-Angle Formulas

Suppose we want to accurately position the Hawaiian Star Compass on the Unit Circle. In Figure 1.1.4,the house for Manu is located at halfway between Hikina and ʻĀkau, resulting in an angle of 45. By applying right triangle trigonometry, we can determine the exact coordinates of Manu as (cos(45),sin(45))=(22,22). However, as we move to the house of ʻĀina, located halfway between Manu and Hikina, we encounter a problem. The angle for ʻĀina is 22.5, which is not explicitly listed listed in Table 1.5.18. Therefore, we must resort to a calculator for numerical approximations.
In this section, we will learn about the double and half-angle formulas for trigonometry. These formulas allow us to determine exact trigonometric function values for angles that are double or half of known values. This will enable us to use our existing knowledge of trigonometric functions at 45 and apply the half-angle formulas to obtain exact values at 22.5.

Subsection 3.3.1 Double-Angle Formulas

Recall the addition formula for sine:
(3.3.1)sin(α+β)=sinαcosβ+cosαsinβ
Consider the case when the two angles are equal. We will call this angle θ, so let α=θ and β=θ. Then Eq (3.3.1) becomes
sin(θ+θ)=sinθcosθ+cosθsinθsin(2θ)=2sinθcosθ
Thus we obtain a formula for sine of twice the angle θ.

Definition 3.3.1. Double-Angle Formulas.

sin2θ=2sinθcosθcos2θ=cos2θsin2θcos2θ=12sin2θcos2θ=2cos2θ1tan2θ=2tanθ1tan2θ
The proofs for the Double-Angle Formulas for Cosine and Tangent are left as exercises (Exercise 3.3.4.6-Exercise 3.3.4.9).

Remark 3.3.2.

Notice that there are three variations of the double-angle formula for cosine. All three equations give the correct answer; however, one version may be more convenient depending on the given information. For example, if you are given the value of sinθ, it may be easier to select the version that solely involves sinθ and does not include cosθ.

Example 3.3.3.

Given sinθ=513 and θ lies in Quadrant III, find the exact value of
(a)
sin(2θ)
Solution.
By the double-angle formula, we have sin(2θ)=2sinθcosθ. We are given the value of sinθ, but we do not have cosθ. To find cosθ, we will draw the triangle formed from sinθ=513 where θ lies in Quadrant III.
Using the Pythagorean Theorem, we can solve the triangle:
x2+(5)2=132x2+25=169x2=144x=12
Thus we have
cosθ=adjacenthypotenuse=1213
and
tanθ=oppositeadjacent=512.
With this new information, we can use the double-angle formula to find sin(2θ):
sin(2θ)=2sinθcosθ=2(513)(1213)=120169
(b)
cos(2θ)
Solution.
To compute cos2θ, notice there are three different formulas: cos2θ=cosθsin2θ, cos2θ=12sin2θ, or cos2θ=2cos2θ1. Using any of the three equations will give us the correct answer. However, given that we know sinθ=513, it may be easier to use cos2θ=12sin2θ, since the other two equations require us to know cosθ.
Without having to draw the triangle, we could get
cos2θ=12sin2θ=12(513)2=12(25169)=16916950169=119169
(c)
tan(2θ)
Solution.
Using the double-angle formula for tangent, we get
tan2θ=2tanθ1tan2θ=2(512)1(512)2=1012125144=1012119144=1012144119=120119

Example 3.3.4.

Write cos(3θ) in terms of sinθ.

Solution.

sin(3θ)=sin(2θ+θ)=sin(2θ)cosθ+cos(2θ)sinθaddition formula=(2sinθcosθ)cosθ+(cos2θsin2θ)sinθdouble-angle formula=2sinθcos2θ+sinθcos2θsin3θ=3sinθcos2θsin3θ=3sinθ(1sin2θ)sin3θPythagorean Identity=3sinθ3sin3θsin3θ=3sinθ4sin3θ

Subsection 3.3.2 Reducing Powers Formulas

You may notice that the double-angle formula for cosine expresses a trigonometric function in terms of the square of another trigonometric function. By rearranging the terms, we can derive formulas for reducing the powers of sine, cosine, and tangent expressions with even powers to terms involving only cosine. These formulas are particularly useful in calculus.

Definition 3.3.5. Formulas for Reducing Powers.

sin2θ=1cos(2θ)2
cos2θ=1+cos(2θ)2
tan2θ=1cos(2θ)1+cos(2θ)

Proof.

To prove the first formula, solve for sin2θ in the double-angle formula: cos2θ=12sin2θ. The second formula is obtained similarly by solving for cos2θ in the formula: cos2θ=2cos2θ1. The first two formula can be used to obtain the third formula:
tan2θ=sin2θcos2θ=1cos(2θ)21+cos(2θ)2=1cos(2θ)1+cos(2θ)

Example 3.3.6.

Write sin4θ as an expression that does not involve powers of sine or cosine greater than 1.

Solution.

We will use the Reducing Powers Formula twice.
sin4θ=(sin2θ)2=(1cos(2θ)2)2reducing powers=14(12cos(2θ)+cos2(2θ))=14(12cos(2θ)+1+cos(4θ)2)reducing powers=14(12cos(2θ)+12+cos(4θ)2)=14(3242cos(2θ)+12cos(4θ))=18(34cos(2θ)+cos(4θ))

Subsection 3.3.3 Half-Angle Formulas

Another set of useful formulas are the half-angle formulas.

Definition 3.3.7. Half-Angle Formulas.

sinθ2=±1cosθ2,cosθ2=±1+cosθ2,tanθ2=±1cosθ1+cosθ
The choice of the + or - sign depends on the Quadrant in which θ/2 lies.

Proof.

We take the square root on both sides of the Formulas for Reducing Powers (Definition 3.3.5) and halve the angle (θ becomes θ2 and 2θ becomes θ) to arrive at our formulas.

Example 3.3.8. Locating ʻĀina.

We are now ready to revisit the problem posed at the start of this section when we were asked to determine the exact coordinates of the house ʻĀina on the Unit Circle.

Solution.

We know the coordinates are at
(cos22.5,sin22.5).
To find the exact value of cos22.5, we will use the half-angle formula:
cos22.5=cos(452)=1+cos452=1+222=22+222=2+24=122+2
Since the half-angle formula has ±, we check the quadrant. In this case, our angle is 22.5, which is in Quadrant I. Therefore, we choose the positive value.
Finding the exact value of sin22.5 is left for Exercise 3.3.4.1.

Example 3.3.9.

Given sinθ=513 and θ lies in Quadrant III, find the exact value of
(a)
sinθ2
Solution.
Notice the Half-Angle Formulas all require us to know cosθ. Since the given information describes the same triangle in Example 3.3.3, we refer to that problem to get cosθ=1213.
Next, since θ is in Quadrant III, 180<θ<270, so dividing by 2 gives us 1802<θ2<2702 or 90<θ2<135. Therefore, we conclude that θ2 lies in Quadrant II.
To calculate sinθ2, we first note that because θ2 lies in Quadrant II, sinθ2>0 so we will choose the positive (+) sign in the Half-Angle Formula:
sinθ2=1cosθ2=1(1213)2=1+12132=1313+12132=25132=2526
(b)
cosθ2
Solution.
Since θ2 is in Quadrant II, we know that cosθ2<0 so we will choose the negative (-) sign in the Half-Angle Formula:
cosθ2=1+cosθ2=1+(1213)2=112132=131312132=1132=126
(c)
tanθ2
Solution.
Since θ2 is in Quadrant II, we know that tanθ2<0 so we will choose the negative (-) sign in the Half-Angle Formula:
tanθ2=1cosθ1+cosθ=1(1213)1+(1213)=1+121311213=1313+121313131213=2513113=2513131=25=5
We can derive another formula for tanθ2 that does not involve ±:

Definition 3.3.10. Half-Angle Formulas for Tangent.

tanθ2=±1cosθ1+cosθ=1cosθsinθ=sinθ1+cosθ

Proof.

We begin by first multiplying both sides of the Sine formula for Reducing Powers by 2 and halving the angle:
1cosθ=2sin2(θ2)
and applying the double-angle formula to:
sinθ=sin2θ2=2sinθ2cosθ2
Dividing the two preceding results
1cosθsinθ=2sin2(θ2)2sinθ2cosθ2=sinθ2cosθ2=tanθ2
Thus
tanθ2=1cosθsinθ
Similarly, it can be shown that
tanθ2=sinθ1+cosθ

Example 3.3.11.

Calculate tanθ2 from Example 3.3.9 using the above formula.

Solution.

tanθ2=1cosθsinθ=1(1213)513=2513513=2513(135)=5
Note: We obtained the same result for tanθ2 as we did in Example 3.3.9. In this example, we did not have to determine if tanθ2 was positive or negative, however, we do need to know the values of both sinθ and cosθ.

Exercises 3.3.4 Exercises

Exercise Group.

The house ʻĀina is located at 22.5=452 and the house Nā Leo is located at 67.5=1352. Use the half-angle formulas to evaluate the exact value of the given expression at each of these houses.

6.

Use the Addition Formula, cos(α+β)=cosαcosβsinαsinβ, to prove the double angle formula for cosine:
cos2θ=cos2θsin2θ.

7.

Use the Pythagorean Identity (sin2θ+cos2θ=1) and the result from Exercise 3.3.4.6 to prove
cos2θ=12sin2θ.

8.

Use the Pythagorean Identity (sin2θ+cos2θ=1) and the result from Exercise 3.3.4.6 to prove
cos2θ=2cos2θ1.

9.

Use the addition formula, tan(α+β)=tanα+tanβ1tanαtanβ, to prove the double angle formula for tangent:
tan2θ=2tanθ1tan2θ.

Exercise Group.

Use the figure below to find the exact values for each of the following exercises.

Exercise Group.

Use the figure below to find the exact values for each of the following exercises.

Exercise Group.

Find the exact value of each expression given cosθ=35 and θ is in Quadrant III.

Exercise Group.

Find the exact value of each expression given sinθ=817 and 270<θ<360.

Exercise Group.

Find the exact value of each expression given cosθ=23 and 3π2<θ<2π.

Exercise Group.

Use the Half-Angle Formula to find the exact value of each of the following

Exercise Group.

Write each of the following as expressions that do not involve powers of sine or cosine greater than 1.
60.
cos4θ
Answer.
38+12cos(2θ)+18cos(4θ)
61.
sin3(2θ)
Answer.
12sin(2θ)(1cos(4θ))
62.
sin2θcos4θ
Answer.
116(1+cos(2θ)cos(4θ)cos(2θ)cos(4θ))
63.
sin4θcos2θ
Answer.
116(1cos(2θ)cos(4θ)+cos(2θ)cos(4θ))

64.

Write sin2θcos2θ expressions that does not involve powers of sine or cosine greater than 1.

Exercise Group.

Use half angle to find the exact deviation for the indicated angle, θ.

Exercise Group.

Verify the identity
68.
(sinθ+cosθ)2=1+sin(2θ)
69.
cos(2θ)=cotθtanθcotθ+tanθ
70.
(sin2θ1)2=cos(2θ)+sin4θ
71.
cos2(3θ)sin2(3θ)=cos(6θ)
72.
cos4θsin4θ=cos(2θ)
73.
sin(6θ)=2sin(3θ)cos(3θ)
74.
cot(2θ)=1tan2θ2tanθ
75.
csc2(θ2)=21cosθ
76.
sec2(θ2)=21+cosθ
77.
2tanθ1+tan2θ=sin(2θ)