Double Angle Identities

Use sum and difference formulas for trigonometric functions. Grade 12 Math 30-1.

Lesson 5.4 of Trigonometric Identities in Math 30-1 — Alberta curriculum lessons.

What you'll learn in this lesson

• The Double Angle Identities & Simplifying. Derive the double angle identities for sine and cosine from the sum identities, learn all equivalent cosine forms, and practice collapsing expressions into a single trig ratio or exact value.
• Application & Given Angle Information. Apply the double angle identities to problems with given quadrant information, unit circle points, finding unknown constants, and amplitude/period/phase shift analysis.

Inside the lesson — a free preview

Where Do They Come From?

If we double an angle of measure α, the new angle will have measure 2α. To develop the double angle identities we use the sum identities from Lesson 5.3.

Deriving 2α. Recall: (α + β) = αβ + αβ

Now suppose β = α:

(α + α) = αα + αα

2α = 2αα

⭐ This is on your formula sheet.

Cosine Double Angle — Formula Sheet Form & Equivalent Forms. Starting from (α + β) = αβ - αβ and setting β = α:

(α + α) = αα - αα

2α = ^2α - ^2α

⭐ This is on your formula sheet. It can be rewritten in two more useful forms using the Pythagorean identity ^2α + ^2α = 1.

Equivalent Form 1 — Substitute ^2α = 1 - ^2α. Starting from the formula sheet form 2α = ^2α - ^2α, replace ^2α using the Pythagorean identity:

2α = ^2α - (1 - ^2α) = 2^2α - 1

Rearranging to isolate ^2α (useful for solving equations):

^2α = (1 + 2α)/(2)

Equivalent Form 2 — Substitute ^2α = 1 - ^2α. Starting from the formula sheet form 2α = ^2α - ^2α, replace ^2α using the Pythagorean identity:

2α = (1 - ^2α) - ^2α = 1 - 2^2α

Rearranging to isolate ^2α (useful for solving equations):

^2α = (1 - 2α)/(2)

📌 Important: The argument for (2α) and (2α) is (2α). The only way to change the argument is by applying one of the double angle identities.

Worked example: Write each expression as a single trig function

Part a) 6(π)/(10)(π)/(10). Formula: 2α = 2αα

6(π)/(10)(π)/(10)

Factor out 3 so the remaining expression matches 2αα with α = (π)/(10):

= 3(2(π)/(10)(π)/(10))

= 3(2 · (π)/(10))

= 3(π)/(5)

Part b) 1 - 2^2(3A). Formula: 2α = 1 - 2^2α (Equivalent Form 2)

1 - 2^2(3A)

The expression already matches the pattern exactly with α = 3A:

= (2 · 3A)

= 6A

Part c) ^2 (52x) - ^2 (52x)2 (52x) (52x). Numerator formula: 2α = ^2α - ^2α (formula sheet form)

Denominator formula: 2α = 2αα (formula sheet form)

Both use α = (5)/(2)x:

^2 (52x) - ^2 (52x)2 (52x) (52x) = (2 · 52x) (2 · 52x) = ( 5x)/( 5x)

= 5x

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