In this section we will include several new identities to the collection we estab- lished in the previous section. These new identities are called “Double-Angle Identities” because they typically deal with relationships between trigonometric functions of a particular angle and functions of “two times” or double the original angle.

To establish the validity of these identities we need to use what are known as the Sum and Difference Identities. These are identities that deal with expressions such as sin(α + β). First we will establish an expression that is equivalent to

cos(α−β).

Let’s start with the unit circle:

(1,0) (cosβ,sinβ) β (cosα,sinα) α α−β

If we rotate everything in this picture clockwise so that the point labeled(cosβ,sinβ)

slides down to the point labeled(1,0), then the angle of rotation in the diagram will beα−βand the corresponding point on the edge of the circle will be:

(cos(α−β),sin(α−β)).

3.2. DOUBLE-ANGLE IDENTITIES 111

(cos(α−β),sin(α−β))

(1,0)

α−β

Since the the second diagram is created by rotating the lines and points from the first diagram, the distance between the points(cosα,sinα)and(cosβ,sinβ)in the first diagram is the same as the distance between(cos(α−β),sin(α−β))and the point(1,0)in the second diagram.

(1,0) (cosβ,sinβ) β d (cosα,sinα) α α−β (cos(α−β),sin(α−β)) (1,0) d α−β

112 CHAPTER 3. TRIGONOMETRIC IDENTITIES AND EQUATIONS

We can represent this distancedwith the distance formula used to calculate the distance between two points in the coordinate plane:

The distance between the points(x1, y1)and(x2, y2)is

d =p(x2−x1)2+ (y2−y1)2

So, in the first diagram the distancedwill be:

d=p(cosα−cosβ)2 _{+ (sin}_{α}_{−}_{sin}_{β}_{)}2

In the second diagram the distancedwill be:

d=p(cos(α−β)−1)2 _{+ (sin(}_{α}_{−}_{β}_{)}_{−}_{0)}2

Since these distances are the same, we can set them equal to each other:

p

(cosα−cosβ)2_{+ (sin}_{α}_{−}_{sin}_{β}_{)}2 _{=}p_{(cos(}_{α}_{−}_{β}_{)}_{−}_{1)}2_{+ (sin(}_{α}_{−}_{β}_{)}_{−}_{0)}2

We’ll square both sides to clear the radicals:

(cosα−cosβ)2_{+ (sin}_{α}_{−}_{sin}_{β}_{)}2 _{= (cos(}_{α}_{−}_{β}_{)}_{−}_{1)}2_{+ (sin(}_{α}_{−}_{β}_{)}_{−}_{0)}2

Next, we’ll rewrite(sin(α−β)−0)2assin2(α−β):

3.2. DOUBLE-ANGLE IDENTITIES 113

Now we’ll work to simplify the expressions on the left-hand side of this equation.

(cosα−cosβ)2_{+ (sin}_{α}_{−}_{sin}_{β}_{)}2 _{= (cos(}_{α}_{−}_{β}_{)}_{−}_{1)}2_{+ sin}2_{(}_{α}_{−}_{β}_{)}

First, each one needs to be squared:

(cosα−cosβ)2 _{= cos}2_{α}_{−}_{2 cos}_{α}_{cos}_{β}_{+ cos}2_{β}

(sinα−sinβ)2 _{= sin}2_{α}_{−}_{2 sin}_{α}_{sin}_{β}_{+ sin}2_{β}

So, the left-hand side will now be:

cos2α−2 cosαcosβ+ cos2β+ sin2α−2 sinαsinβ+ sin2β

If we rearrange this a little, it will simplify nicely:

cos2_{α}_{−}_{2 cos}_{α}_{cos}_{β}_{+ cos}2_{β}_{+ sin}2_{α}_{−}_{2 sin}_{α}_{sin}_{β}_{+ sin}2_{β}

sin2α+ cos2_{α}_{+ sin}2_{β}_{+ cos}2_{β}_{−}_{2 cos}_{α}_{cos}_{β}_{−}_{2 sin}_{α}_{sin}_{β}

Notice the Pythagorean Identities at the front of this expression - these are each equal to1:

sin2α+ cos2α+ sin2β+ cos2β−2 cosαcosβ−2 sinαsinβ

1 + 1−2 cosαcosβ−2 sinαsinβ

2−2 cosαcosβ−2 sinαsinβ

114 CHAPTER 3. TRIGONOMETRIC IDENTITIES AND EQUATIONS

Now that we’ve simplified the left-hand side, we’ll simplify the right-hand side. First we’ll square the expression(cos(α−β)−1)2:

(cos(α−β)−1)2 = cos2(α−β)−2 cos(α−β) + 1

So, the right-hand side is now:

cos2_{(}_{α}_{−}_{β}_{)}_{−}_{2 cos(}_{α}_{−}_{β}_{) + 1 + sin}2_{(}_{α}_{−}_{β}_{)}

If we rearrange this expression, we’ll again have a nice Pythagorean Identity:

sin2(α−β) + cos2_{(}_{α}_{−}_{β}_{)}_{−}_{2 cos(}_{α}_{−}_{β}_{) + 1}

1−2 cos(α−β) + 1 2−2 cos(α−β) 2(1−cos(α−β))

So the left-hand side was equal to:

2(1−cosαcosβ−sinαsinβ)

And the right-hand side was equal to:

2(1−cos(α−β))

So, our original statement in simplified form is:

3.2. DOUBLE-ANGLE IDENTITIES 115

If we divide by2on both sides, we’ll have:

1−cosαcosβ−sinαsinβ = 1−cos(α−β)

then subtract1

−cosαcosβ−sinαsinβ =−cos(α−β)

and multiply through by−1 cosαcosβ+ sinαsinβ = cos(α−β)

So,cos(α−β) = cosαcosβ+ sinαsinβ.

This will help us to generate the double-angle formulas, but to do this, we don’t wantcos(α−β), we wantcos(α+β)(you’ll see why in a minute).

So, to change this around, we’ll use identities for negative angles. Recall that in the fourth quadrant the sine function is negative and the cosine function is positive. For this reason,sin(−θ) =−sin(θ)andcos(−θ) = cos(θ).

Now we can say that cos(2θ) = cos(θ +θ) = cos(θ −(−θ)). Going back to our identity forcos(α−β), we can say that:

cos(θ−(−θ)) = cosθcos(−θ) + sinθsin(−θ) cos(θ−(−θ)) = cosθcosθ+ sinθ(−sinθ) cos(θ−(−θ)) = cosθcosθ−sinθsinθ

cos(θ−(−θ)) = cos2θ−sin2θ

cos(θ+θ) = cos2θ−sin2θ

cos(2θ) = cos2θ−sin2θ

This is the double-angle identity for the cosine: cos(2θ) = cos2_{θ}_{−}_{sin}2_{θ}

116 CHAPTER 3. TRIGONOMETRIC IDENTITIES AND EQUATIONS

This identity actually appears in any one of three forms because the Pythagorean Identities can be applied to this to change its appearance:

cos(2θ) = cos2θ−sin2θ

cos(2θ) = 1−sin2θ−sin2θ

cos(2θ) = 1−2 sin2θ

If we substitute for thesin2θterm:

cos(2θ) = cos2θ−sin2θ

cos(2θ) = cos2θ−(1−cos2θ) cos(2θ) = cos2θ−1 + cos2θ

cos(2θ) = 2 cos2θ−1

So, the three forms of the cosine double angle identity are:

cos(2θ) = cos2θ−sin2θ

= 2 cos2θ−1 = 1−2 sin2θ

The double-angle identity for the sine function uses what is known as the co- function identity. Remember that, in a right triangle, the sine of one angle is the same as the cosine of its complement (which is the other acute angle). This is because the adjacent side for one angle is the opposite side for the other angle. The denominator in both cases is the hypotenuse, so the cofunctions of comple- mentary angles are equal.

3.2. DOUBLE-ANGLE IDENTITIES 117

In the diagram below, we can see this more clearly:

θ

a

b c

90◦−θ

In the diagram above note that:

sinθ =a_{c}= cos(90◦ −θ)

So, if we want an identity forsin(θ+θ), we’ll start withsin(α+β)which is equiv- alent tocos(90◦−(α+β)). We’ll use a trick here and restate this as:

sin(α+β) = cos(90◦−(α+β)) = cos(90◦−α−β) = cos((90◦−α)−β)

= cos(90◦−α) cosβ+ sin(90◦−α) sinβ

= sinαcosβ+ cosαsinβ

Now, we can use this to find an expression forsin 2θ= sin(θ+θ):

sin 2θ = sin(θ+θ)

= sinθcosθ+ cosθsinθ

118 CHAPTER 3. TRIGONOMETRIC IDENTITIES AND EQUATIONS

Here is a summary of all the identities we’ve worked with in this chapter:

Pythagorean Identities Reciprocal Identities

sin2θ+ cos2_{θ} _{= 1} _{tan}_{θ} _{=}sinθ

cosθ

tan2_{θ}_{+ 1 = sec}2_{θ} _{cot}_{θ} _{=}cosθ

sinθ

1 + cot2θ= csc2θ secθ =_{cos}1_{θ}
cscθ =_{sin}1_{θ}

Double-Angle Identities

cos(2θ) = cos2_{θ}_{−}_{sin}2_{θ}

cos(2θ) = 2 cos2θ−1 cos(2θ) = 1−2 sin2θ

sin(2θ) = 2 sinθcosθ

Working problems involving double-angle identities is very similar to the other identities we’ve worked with previously - you just have more identities to choose from!

**Example**

Verify the given identity:cos 2x=1_{1+tan}−tan22x_{x}

We have three possible identities to choose from for the left-hand side, so we’ll wait on that for a moment while we simplify the right-hand side.

3.2. DOUBLE-ANGLE IDENTITIES 119
1−tan2x
1 + tan2_{x} =
1−tan2x
sec2_{x}
= 1
sec2_{x}−
tan2_{x}
sec2_{x}
= cos2x−(
sin2x
cos2_{x})
( 1
cos2_{x})
= cos2x− sin
2_{x}
cos2_{x} ·
cos2_{x}
1
= cos2x− sin
2_{x}
cos2x
·cos2x
1
= cos2x−sin2x

This is one of the identities forcos(2θ)so we can stop and simply statecos(2x) =
cos2_{x}_{−}_{sin}2_{x}

120 CHAPTER 3. TRIGONOMETRIC IDENTITIES AND EQUATIONS

**Exercises 3.2**

In each problem verify the given trigonometric identity.

1. _{cos}2 sin2_{x}x_{−}cos_{sin}2x_{x}= tan(2x) 2. sec(2x) =

sec2x

2−sec2_{x}

3. sin(2x) cscx= 2 cosx 4. _{sin(2}2 cos_{x}x_{)}= cscx

5. cos(2_{sin}_{x}x)+ sinx=cot_{sec}_{x}x 6. sin_{sec}x+sin(2_{x}_{+2}x)= sinxcosx

7. (sinx+ cosx)2 _{= 1 + sin(2}_{x}_{)} _{8.} _{(sin}2_{x}_{−}_{1)}2 _{= sin}4_{x}_{+ cos(2}_{x}_{)}

9. 2 cosx−cos(2_{cos}_{x}x)= secx 10. 1+cos(2_{1}_{−}_{cos(2}x_{x})_{)}= cot2x

11. cos(2_{sin}2x_{x})= cot2x−1 12.

cos(2x) sin2x = csc

2_{x}_{−}_{2}

13. cot_{cot}x_{x}−_{+tan}tanx_{x}= cos 2x 14. sin 2x=2(tanx−cotx)

tan2x−cot2x

15. 2 cos 2_{sin 2}_{x}x= cotx−tanx 16. tan 2x=_{cot}_{x}_{−}2_{tan}_{x}

17. _{1+cos}sinx_{x}+1+cos_{sin}_{x}x= 2 cscx 18. tanx+ cotx= 2 csc(2x)

19. cos(2x) =cot2x−1 cot2x+1 20. sin(2x) = 2 tanx 1+tan2x 21. 2 sin 2 x

sin(2x)+ cotx= secxcscx 22. sec2xcos(2x) = sec2x−2 tan2x