# 3. Double-Angle Formulas

by M. Bourne

The double-angle formulas can be quite useful when we need to simplify complicated trigonometric expressions later.

With these formulas, it is better to remember where they come from, rather than trying to remember the actual formulas. In this way, you will understand it better and have less to clutter your memory with.

## Sine of a Double Angle

sin 2

α= 2 sinαcosα

### Proof

Recall from the last section, the sine of the sum of two angles:

sin(

+α) = sinβαcos+ cosβαsinβ

We will use this to obtain the sine of a double angle.

If we take the left hand side (LHS):

sin(

α+)β

and replace * β* with

*α*, we get:

sin(

α+β) = sin(α+α) = sin 2α

Consider the RHS:

sin

αcosβ+ cosαsinβ

Since we replaced *β* in the LHS with *α*, we need to do the same on the right side. We do this, and obtain:

sin

αcosα+ cosαsinα= 2 sinαcosα

Putting our results for the LHS and RHS together, we obtain the important result:

sin 2

α= 2 sinαcosα

This result is called the **sine of a double angle. ** It is useful for simplifying expressions later.

Continues below ⇩

## Cosine of a Double Angle

Using a similar process, we obtain the **cosine of a double angle** formula:

cos 2

α= cos^{2}α− sin^{2}α

### Proof

This time we start with the cosine of the sum of two angles:

cos(

α+β) = cosαcosβ− sinαsinβ,

and once again replace *β* with *α* on both the LHS and RHS, as follows:

LHS = cos(

α+α) = cos(2α)RHS = cos

αcosα− sinαsinα= cos^{2}α− sin^{2}α

## Different forms of the Cosine Double Angle Result

By using the result sin^{2} *α* + cos^{2} *α* = 1, (which we found in Trigonometric Identities) we can write the RHS of the above formula as:

cos^{2} *α* − sin^{2} *α*

= (1− sin

^{2}α) − sin^{2}α= 1− 2sin

^{2}α

Likewise, we can substitute (1 − cos ^{2} *α*) for sin^{2} *α* into our RHS and obtain:

cos^{2} *α* − sin^{2} *α*

= cos

^{2}α− (1 − cos^{2}α)= 2cos

^{2}α− 1

### Summary - Cosine of a Double Angle

The following have equivalent value, and we can use whichever one we like, depending on the situation:

cos 2α = cos

^{2}α − sin^{2}αcos 2α = 1− 2 sin

^{2}αcos 2α = 2 cos

^{2}α − 1

### Example 1

Find cos 60° by using the functions of 30°.

### Example 2

Find the exact value of `cos 2x` if `sin x= -12/13` (in Quadrant III).

### Exercises

1. Without finding `x`, find the exact value of `sin 2x`* *if `cos x= 4/5` (in Quadrant I).

2. Prove that

`(1-tan^2x)/(sec^2x)=cos2x`

3. Prove that

`2 csc 2x tan x = sec^2x`

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