Multiple-Angle Identities

Multiple-angle identities are trigonometric formulas that connect the sine, cosine, and tangent of an angle with the trigonometric functions of its multiples or fractions.

For example, if an angle is θ, we may want to calculate:

• sin 2θ — the sine of twice the angle.

• cos 3θ — the cosine of three times the angle.

• tan 2θ — the tangent of twice the angle.

• sin(θ/2) — the sine of half the angle.

These identities allow us to simplify expressions, calculate exact values, derive new mathematical relationships, and solve equations involving several angles.

Multiple-angle identities are particularly important in advanced trigonometry, calculus, wave motion, electrical engineering, and the study of oscillations.

Understanding Multiple Angles

A multiple angle is formed by multiplying an angle by a number.

For example, if θ = 30°:

The values are different because changing the angle changes the position on the unit circle.

Important Notation

The expressions sin 2θ and sin² θ are not the same.

sin 2θ = sin(2θ)

sin² θ = (sin θ)²

Similarly:

cos 3θ = cos(3θ)

cos³ θ = (cos θ)³

This distinction is essential when applying multiple-angle identities.

Reviewing Compound Angle Identities

Multiple-angle identities are derived from the sum and difference identities.

The most important compound angle identities are:

sin(A + B) = sin A cos B + cos A sin B

sin(A − B) = sin A cos B − cos A sin B

cos(A + B) = cos A cos B − sin A sin B

cos(A − B) = cos A cos B + sin A sin B

For tangent:

tan(A + B) = (tan A + tan B)/(1 − tan A tan B)

tan(A − B) = (tan A − tan B)/(1 + tan A tan B)

The tangent formulas require tan A and tan B to be defined and their denominators to be nonzero.

To derive a double-angle identity, we set:

A = θ and B = θ

To derive a triple-angle identity, we can write:

3θ = 2θ + θ

Half-angle identities can be derived by rearranging double-angle identities.

Deriving the Sine Double-Angle Identity

Start with the sine addition identity:

sin(A + B) = sin A cos B + cos A sin B

Let:

A = θ

B = θ

Then:

sin(θ + θ) = sin θ cos θ + cos θ sin θ

Combine like terms:

sin 2θ = 2sin θ cos θ

Therefore:

sin 2θ = 2sin θ cos θ

Worked Example: Finding sin 60°

Let θ = 30°.

sin 60° = 2sin 30° cos 30°

Substitute:

= 2(1/2)(√3/2)

Therefore:

sin 60° = √3/2

Worked Example: Finding sin 120°

Let θ = 60°.

sin 120° = 2sin 60° cos 60°

= 2(√3/2)(1/2)

Therefore:

sin 120° = √3/2

The result is positive because 120° lies in Quadrant II.

Deriving the Cosine Double-Angle Identities

Start with the cosine addition identity:

cos(A + B) = cos A cos B − sin A sin B

Let:

A = B = θ

Then:

cos 2θ = cos² θ − sin² θ

Therefore:

cos 2θ = cos² θ − sin² θ

This is the first form of the cosine double-angle identity.

Deriving the Second Form

Use the Pythagorean identity:

sin² θ + cos² θ = 1

Rearrange:

sin² θ = 1 − cos² θ

Substitute:

cos 2θ = cos² θ − (1 − cos² θ)

Expand:

= cos² θ − 1 + cos² θ

Simplify:

cos 2θ = 2cos² θ − 1

Deriving the Third Form

Start with:

cos 2θ = cos² θ − sin² θ

Use:

cos² θ = 1 − sin² θ

Substitute:

cos 2θ = (1 − sin² θ) − sin² θ

Simplify:

cos 2θ = 1 − 2sin² θ

Summary

All three forms are equivalent:

cos 2θ = cos² θ − sin² θ

cos 2θ = 2cos² θ − 1

cos 2θ = 1 − 2sin² θ

Choosing the Correct Form

Deriving the Tangent Double-Angle Identity

Start with:

tan(A + B) = (tan A + tan B)/(1 − tan A tan B)

Let:

A = B = θ

Then:

tan 2θ = (tan θ + tan θ)/(1 − tan² θ)

Simplify:

tan 2θ = 2tan θ/(1 − tan² θ)

This quotient form requires tan θ to be defined and 1 − tan² θ ≠ 0.

Worked Example

Suppose:

tan θ = 1/3

Find tan 2θ.

Use:

tan 2θ = 2tan θ/(1 − tan² θ)

Substitute:

= 2(1/3)/(1 − (1/3)²)

= (2/3)/(1 − 1/9)

= (2/3)/(8/9)

= (2/3)(9/8)

Therefore:

tan 2θ = 3/4

Summary of Double-Angle Identities

These identities can be applied in either direction.

For example:

2sin θ cos θ = sin 2θ

and:

1 − 2sin² θ = cos 2θ

Recognising these patterns is especially useful when simplifying expressions.

Deriving the Sine Triple-Angle Identity

A triple angle is three times an angle.

To derive sin 3θ, write:

3θ = 2θ + θ

Start with:

sin(A + B) = sin A cos B + cos A sin B

Let:

A = 2θ

B = θ

Then:

sin 3θ = sin 2θ cos θ + cos 2θ sin θ

Substitute:

sin 2θ = 2sin θ cos θ

cos 2θ = 1 − 2sin² θ

Therefore:

sin 3θ = (2sin θ cos θ)cos θ + (1 − 2sin² θ)sin θ

Simplify:

= 2sin θ cos² θ + sin θ − 2sin³ θ

Use:

cos² θ = 1 − sin² θ

Substitute:

= 2sin θ(1 − sin² θ) + sin θ − 2sin³ θ

Expand:

= 2sin θ − 2sin³ θ + sin θ − 2sin³ θ

Combine like terms:

sin 3θ = 3sin θ − 4sin³ θ

This is the sine triple-angle identity.

Alternative Form

Factor sin θ:

sin 3θ = sin θ(3 − 4sin² θ)

Both forms are useful.

Applying the Sine Triple-Angle Identity

Worked Example: Finding sin 90°

Let:

θ = 30°

Then:

3θ = 90°

Use:

sin 3θ = 3sin θ − 4sin³ θ

Substitute:

sin 90° = 3sin 30° − 4sin³ 30°

= 3(1/2) − 4(1/2)³

= 3/2 − 4(1/8)

= 3/2 − 1/2

Therefore:

sin 90° = 1

Worked Example: Finding sin 180°

Let:

θ = 60°

Then:

sin 180° = 3sin 60° − 4sin³ 60°

Substitute:

= 3(√3/2) − 4(√3/2)³

Since:

(√3/2)³ = 3√3/8

Then:

= 3√3/2 − 4(3√3/8)

= 3√3/2 − 3√3/2

Therefore:

sin 180° = 0

Deriving the Cosine Triple-Angle Identity

Start with:

cos 3θ = cos(2θ + θ)

Apply the cosine addition identity:

cos 3θ = cos 2θ cos θ − sin 2θ sin θ

Substitute:

cos 2θ = 2cos² θ − 1

sin 2θ = 2sin θ cos θ

Then:

cos 3θ = (2cos² θ − 1)cos θ − (2sin θ cos θ)sin θ

Expand:

= 2cos³ θ − cos θ − 2sin² θ cos θ

Use:

sin² θ = 1 − cos² θ

Substitute:

= 2cos³ θ − cos θ − 2(1 − cos² θ)cos θ

Expand:

= 2cos³ θ − cos θ − 2cos θ + 2cos³ θ

Combine like terms:

cos 3θ = 4cos³ θ − 3cos θ

Alternative Form

Factor cos θ:

cos 3θ = cos θ(4cos² θ − 3)

Applying the Cosine Triple-Angle Identity

Worked Example: Finding cos 180°

Let:

θ = 60°

Then:

3θ = 180°

Use:

cos 3θ = 4cos³ θ − 3cos θ

Substitute:

cos 180° = 4cos³ 60° − 3cos 60°

= 4(1/2)³ − 3(1/2)

= 4(1/8) − 3/2

= 1/2 − 3/2

Therefore:

cos 180° = −1

Worked Example: Finding cos 135°

Let:

θ = 45°

Then:

3θ = 135°

Use:

cos 135° = 4cos³ 45° − 3cos 45°

Substitute:

= 4(√2/2)³ − 3(√2/2)

Since:

(√2/2)³ = √2/4

Then:

= 4(√2/4) − 3√2/2

= √2 − 3√2/2

Therefore:

cos 135° = −√2/2

Deriving the Tangent Triple-Angle Identity

Start with:

tan 3θ = tan(2θ + θ)

Apply the tangent addition identity:

tan 3θ = (tan 2θ + tan θ)/(1 − tan 2θ tan θ)

Let:

t = tan θ

Then:

tan 2θ = 2t/(1 − t²)

Substitute:

tan 3θ = [2t/(1 − t²) + t]/[1 − (2t/(1 − t²))t]

Simplify the numerator:

2t/(1 − t²) + t

= [2t + t(1 − t²)]/(1 − t²)

= (3t − t³)/(1 − t²)

Simplify the denominator:

1 − 2t²/(1 − t²)

= (1 − 3t²)/(1 − t²)

Divide the two expressions:

tan 3θ = [(3t − t³)/(1 − t²)]/[(1 − 3t²)/(1 − t²)]

Cancel the common factor:

tan 3θ = (3t − t³)/(1 − 3t²)

Substitute t = tan θ:

tan 3θ = (3tan θ − tan³ θ)/(1 − 3tan² θ)

This form is valid where tan θ is defined and 1 − 3tan² θ ≠ 0.

The derivation above assumes tan 2θ is defined at the intermediate step. The final formula also holds at angles where tan 2θ is undefined, provided the final formula is defined.

Applying the Tangent Triple-Angle Identity

Worked Example

Suppose:

tan θ = 1/2

Find tan 3θ.

Use:

tan 3θ = (3tan θ − tan³ θ)/(1 − 3tan² θ)

Substitute:

= [3(1/2) − (1/2)³]/[1 − 3(1/2)²]

Simplify the numerator:

= (3/2 − 1/8)/(1 − 3/4)

= (12/8 − 1/8)/(1/4)

= (11/8)/(1/4)

Therefore:

tan 3θ = 11/2

Summary of Triple-Angle Identities

The sine and cosine identities are defined for every real angle.

The tangent identity requires its component expressions and denominator to be defined.

Deriving the Half-Angle Identities

Half-angle identities are obtained by rearranging the cosine double-angle identities.

Deriving the Sine Half-Angle Identity

Start with:

cos 2θ = 1 − 2sin² θ

Rearrange:

2sin² θ = 1 − cos 2θ

Divide by 2:

sin² θ = (1 − cos 2θ)/2

Replace θ with θ/2:

sin²(θ/2) = (1 − cos θ)/2

Take the square root:

sin(θ/2) = ±√[(1 − cos θ)/2]

The sign depends on the quadrant containing θ/2.

Deriving the Cosine Half-Angle Identity

Start with:

cos 2θ = 2cos² θ − 1

Rearrange:

2cos² θ = 1 + cos 2θ

Divide by 2:

cos² θ = (1 + cos 2θ)/2

Replace θ with θ/2:

cos²(θ/2) = (1 + cos θ)/2

Take the square root:

cos(θ/2) = ±√[(1 + cos θ)/2]

Again, the correct sign depends on the quadrant of θ/2.

Deriving the Tangent Half-Angle Identities

Tangent is defined as:

tan(θ/2) = sin(θ/2)/cos(θ/2)

Using the sine and cosine half-angle identities gives:

tan(θ/2) = ±√[(1 − cos θ)/(1 + cos θ)]

where the expression is defined.

Two additional forms are often more convenient.

Start with:

sin θ = 2sin(θ/2)cos(θ/2)

and:

1 + cos θ = 2cos²(θ/2)

Divide:

sin θ/(1 + cos θ)

= [2sin(θ/2)cos(θ/2)]/[2cos²(θ/2)]

Simplify:

= sin(θ/2)/cos(θ/2)

Therefore:

tan(θ/2) = sin θ/(1 + cos θ)

Similarly:

tan(θ/2) = (1 − cos θ)/sin θ

The first rational form is valid when 1 + cos θ ≠ 0. The second requires sin θ ≠ 0. The square-root form requires 1 + cos θ ≠ 0, with the sign chosen from the quadrant.

Summary of Half-Angle Identities

The square-root identities require careful attention to signs.

Applying Half-Angle Identities to Exact Values

Worked Example: Finding sin 15°

Since:

15° = 30°/2

Use:

sin(θ/2) = ±√[(1 − cos θ)/2]

Let:

θ = 30°

Then:

sin 15° = √[(1 − cos 30°)/2]

The positive square root is used because 15° lies in Quadrant I.

Substitute:

= √[(1 − √3/2)/2]

Simplify:

= √[(2 − √3)/4]

Therefore:

sin 15° = √(2 − √3)/2

This is equivalent to:

sin 15° = (√6 − √2)/4

Worked Example: Finding cos 22.5°

Since:

22.5° = 45°/2

Use:

cos(θ/2) = ±√[(1 + cos θ)/2]

Let:

θ = 45°

Then:

cos 22.5° = √[(1 + √2/2)/2]

Simplify:

= √[(2 + √2)/4]

Therefore:

cos 22.5° = √(2 + √2)/2

Determining the Correct Sign in Half-Angle Problems

Half-angle identities contain a ± sign because taking the square root of a squared expression produces two possible signs.

The quadrant of the half angle determines which sign is correct.

Worked Example

Suppose:

cos θ = −3/5

and:

90° < θ < 180°

Find sin(θ/2) and cos(θ/2).

Since θ lies in Quadrant II:

45° < θ/2 < 90°

Therefore, θ/2 lies in Quadrant I.

Both sine and cosine are positive.

Find Sine

sin(θ/2) = √[(1 − cos θ)/2]

Substitute:

= √[(1 − (−3/5))/2]

= √[(8/5)/2]

= √(4/5)

Therefore:

sin(θ/2) = 2√5/5

Find Cosine

cos(θ/2) = √[(1 + cos θ)/2]

Substitute:

= √[(1 − 3/5)/2]

= √(1/5)

Therefore:

cos(θ/2) = √5/5

Power-Reduction Identities

The cosine double-angle identities can be rearranged to express squared sine and cosine functions using double angles.

Start with:

cos 2θ = 1 − 2sin² θ

Rearrange:

sin² θ = (1 − cos 2θ)/2

Similarly:

cos 2θ = 2cos² θ − 1

Rearrange:

cos² θ = (1 + cos 2θ)/2

These are called power-reduction identities.

Why Are They Useful?

Power-reduction identities are useful when:

• Simplifying expressions containing squared trigonometric functions.

• Deriving equivalent identities.

• Calculating average values of periodic functions.

• Integrating trigonometric functions in calculus.

Worked Example

Simplify:

2sin² θ − 1

Use:

cos 2θ = 1 − 2sin² θ

Therefore:

2sin² θ − 1 = −cos 2θ

Worked Example

Rewrite:

cos² θ

using a double angle.

Apply:

cos² θ = (1 + cos 2θ)/2

Deriving Equivalent Forms of Multiple-Angle Identities

Multiple-angle identities can often be rearranged into different forms.

Example: Rewriting the Sine Triple-Angle Identity

Start with:

sin 3θ = 3sin θ − 4sin³ θ

Factor sin θ:

sin 3θ = sin θ(3 − 4sin² θ)

Use:

sin² θ = 1 − cos² θ

Substitute:

= sin θ[3 − 4(1 − cos² θ)]

Simplify:

= sin θ(4cos² θ − 1)

Therefore:

sin 3θ = sin θ(4cos² θ − 1)

This gives another useful form of the sine triple-angle identity.

Example: Rewriting the Cosine Triple-Angle Identity

Start with:

cos 3θ = 4cos³ θ − 3cos θ

Factor cos θ:

cos 3θ = cos θ(4cos² θ − 3)

Use:

cos² θ = 1 − sin² θ

Then:

cos 3θ = cos θ[4(1 − sin² θ) − 3]

Simplify:

cos 3θ = cos θ(1 − 4sin² θ)

These equivalent forms are useful when different trigonometric values are known.

Using Multiple-Angle Identities to Simplify Expressions

Worked Example

Simplify:

6sin θ cos θ

Use:

sin 2θ = 2sin θ cos θ

Therefore:

6sin θ cos θ = 3(2sin θ cos θ)

= 3sin 2θ

Worked Example

Simplify:

4cos³ θ − 3cos θ

Recognise the cosine triple-angle identity:

= cos 3θ

Worked Example

Simplify:

3sin θ − 4sin³ θ

Recognise the sine triple-angle identity:

= sin 3θ

Worked Example

Simplify:

1 − 2sin² θ

Recognise the cosine double-angle identity:

= cos 2θ

Combining Multiple-Angle Identities

Some expressions require more than one identity.

Worked Example

Simplify:

sin 3θ − sin θ

Use:

sin 3θ = 3sin θ − 4sin³ θ

Substitute:

= 3sin θ − 4sin³ θ − sin θ

Combine:

= 2sin θ − 4sin³ θ

Factor:

= 2sin θ(1 − 2sin² θ)

Use:

1 − 2sin² θ = cos 2θ

Therefore:

sin 3θ − sin θ = 2sin θ cos 2θ

Worked Example

Simplify:

cos 3θ + cos θ

Use:

cos 3θ = 4cos³ θ − 3cos θ

Substitute:

= 4cos³ θ − 3cos θ + cos θ

= 4cos³ θ − 2cos θ

Factor:

= 2cos θ(2cos² θ − 1)

Use:

2cos² θ − 1 = cos 2θ

Therefore:

cos 3θ + cos θ = 2cos θ cos 2θ

Verifying Multiple-Angle Identities

A trigonometric identity proof demonstrates that two expressions are equivalent wherever both sides are defined.

Worked Example

Verify:

sin 3θ = sin θ(4cos² θ − 1)

Start with the right-hand side:

sin θ(4cos² θ − 1)

Use:

cos² θ = 1 − sin² θ

Substitute:

= sin θ[4(1 − sin² θ) − 1]

Simplify:

= sin θ(3 − 4sin² θ)

Expand:

= 3sin θ − 4sin³ θ

Use the sine triple-angle identity:

= sin 3θ

Therefore, the identity is verified.

Worked Example

Verify:

cos 3θ = cos θ(1 − 4sin² θ)

Start with:

cos θ(1 − 4sin² θ)

Use:

sin² θ = 1 − cos² θ

Substitute:

= cos θ[1 − 4(1 − cos² θ)]

Simplify:

= cos θ(4cos² θ − 3)

Expand:

= 4cos³ θ − 3cos θ

Use the cosine triple-angle identity:

= cos 3θ

The identity is verified.

Solving Equations Using Double-Angle Identities

Worked Example

Solve:

sin 2x = sin x

for:

0° ≤ x < 360°

Use:

sin 2x = 2sin x cos x

Substitute:

2sin x cos x = sin x

Move everything to one side:

2sin x cos x − sin x = 0

Factor:

sin x(2cos x − 1) = 0

Therefore:

sin x = 0

or:

cos x = 1/2

First Case

sin x = 0

x = 0°, 180°

Second Case

cos x = 1/2

x = 60°, 300°

Final Answer

x = 0°, 60°, 180°, 300°

Solving Equations Using Triple-Angle Identities

Worked Example

Solve:

sin 3x = sin x

for:

0° ≤ x < 360°

Use:

sin 3x = 3sin x − 4sin³ x

Substitute:

3sin x − 4sin³ x = sin x

Rearrange:

2sin x − 4sin³ x = 0

Factor:

2sin x(1 − 2sin² x) = 0

Therefore:

sin x = 0

or:

1 − 2sin² x = 0

First Case

sin x = 0

x = 0°, 180°

Second Case

1 − 2sin² x = 0

sin² x = 1/2

Therefore:

sin x = ±√2/2

The solutions are:

x = 45°, 135°, 225°, 315°

Final Answer

x = 0°, 45°, 135°, 180°, 225°, 315°

Solving Cosine Triple-Angle Equations

Worked Example

Solve:

cos 3x = cos x

for:

0° ≤ x < 360°

Use:

cos 3x = 4cos³ x − 3cos x

Substitute:

4cos³ x − 3cos x = cos x

Rearrange:

4cos³ x − 4cos x = 0

Factor:

4cos x(cos² x − 1) = 0

Therefore:

cos x = 0

or:

cos² x = 1

First Case

cos x = 0

x = 90°, 270°

Second Case

cos² x = 1

cos x = ±1

Therefore:

x = 0°, 180°

Final Answer

x = 0°, 90°, 180°, 270°

Solving Equations Involving Multiple Angles Directly

Sometimes it is easier to solve the equation without expanding the multiple-angle identity.

Worked Example

Solve:

sin 3x = 1/2

for:

0° ≤ x < 360°

Since:

0° ≤ x < 360°

then:

0° ≤ 3x < 1080°

Sine equals 1/2 at:

3x = 30°, 150°, 390°, 510°, 750°, 870°

Divide each angle by 3:

x = 10°, 50°, 130°, 170°, 250°, 290°

Final Answer

x = 10°, 50°, 130°, 170°, 250°, 290°

Why Are There Six Solutions?

The function sin 3x completes three full cycles between 0° and 360°.

Each cycle contains two solutions for sin 3x = 1/2.

Therefore, there are six solutions.

Solving Multi-Step Multiple-Angle Equations

Worked Example

Solve:

2cos 2x − 1 = 0

for:

0° ≤ x < 360°

First isolate cosine:

2cos 2x = 1

cos 2x = 1/2

Since:

0° ≤ 2x < 720°

the solutions for 2x are:

60°, 300°, 420°, 660°

Divide by 2:

x = 30°, 150°, 210°, 330°

Verification

For x = 30°:

2cos 60° − 1 = 2(1/2) − 1 = 0

For x = 150°:

2cos 300° − 1 = 0

For x = 210°:

2cos 420° − 1 = 0

For x = 330°:

2cos 660° − 1 = 0

All four solutions satisfy the equation.

Applications in Physics: Waves and Oscillations

Multiple-angle identities are useful for analysing periodic motion.

A simple sinusoidal wave can be represented by:

y(t) = A sin(ωt)

where:

• A is amplitude.

• ω is angular frequency.

• t is time.

A wave with twice the angular frequency is:

y(t) = A sin(2ωt)

Using the sine double-angle identity:

sin(2ωt) = 2sin(ωt)cos(ωt)

Therefore:

y(t) = 2A sin(ωt)cos(ωt)

A wave with three times the angular frequency is:

y(t) = A sin(3ωt)

Using the triple-angle identity:

y(t) = A[3sin(ωt) − 4sin³(ωt)]

These relationships are useful when analysing higher-frequency components of periodic signals.

Frequency and Period

For:

y(t) = A sin(ωt)

the period is:

P = 2π/ω

For:

y(t) = A sin(2ωt)

the period is:

P = π/ω

For:

y(t) = A sin(3ωt)

the period is:

P = 2π/(3ω)

Therefore, increasing the angular frequency decreases the period.

Applications in Calculus

Multiple-angle identities are especially useful when integrating trigonometric functions.

For example:

∫ sin² x dx

Use the power-reduction identity:

sin² x = (1 − cos 2x)/2

Substitute:

∫ sin² x dx

= ∫(1 − cos 2x)/2 dx

Separate:

= (1/2)∫1 dx − (1/2)∫cos 2x dx

Integrate:

= x/2 − sin 2x/4 + C

Therefore:

∫ sin² x dx = x/2 − sin 2x/4 + C

Similarly:

∫ cos² x dx = x/2 + sin 2x/4 + C

These results are used in calculations involving wave energy, oscillations, and periodic systems.

Applications in Geometry

Half-angle identities allow exact values to be calculated for angles that are difficult to evaluate directly.

Worked Example

A right triangle has:

• Hypotenuse = 12 cm

• Acute angle = 22.5°

Find the exact length of the side opposite the angle.

Use:

opposite = hypotenuse × sin θ

Therefore:

a = 12sin 22.5°

Use the half-angle identity:

sin 22.5° = √(2 − √2)/2

Substitute:

a = 12[√(2 − √2)/2]

Simplify:

a = 6√(2 − √2) cm

Approximately:

a ≈ 4.59 cm

Choosing the Correct Multiple-Angle Identity

A Strategy for Solving Multiple-Angle Problems

When solving problems involving multiple angles:

1. Identify whether the expression contains a double, triple, or half angle.

2. Determine which trigonometric function is involved.

3. Select the most useful identity.

4. Consider whether expanding or factoring will simplify the problem.

5. Use Pythagorean identities to derive equivalent forms when necessary.

6. Substitute known exact values.

7. Check quadrant signs, especially for half-angle formulas.

8. When solving equations, consider whether direct angle solving is easier than expansion.

9. Adjust the interval when the variable is multiplied by 2 or 3.

10. Avoid dividing by expressions that may equal zero.

11. Verify proposed solutions using the original equation.

12. State exact answers whenever possible.

Common Mistakes

Confusing sin 2θ with sin² θ

These expressions are not equivalent.

Confusing sin 3θ with sin³ θ

sin 3θ means sin(3θ), while sin³ θ means (sin θ)³.

Forgetting the coefficient in the sine double-angle identity

The correct identity is sin 2θ = 2sin θ cos θ.

Using the wrong cosine double-angle form

Choose the form that matches the known information.

Confusing triple-angle coefficients

The correct sine triple-angle identity is:

sin 3θ = 3sin θ − 4sin³ θ

The correct cosine triple-angle identity is:

cos 3θ = 4cos³ θ − 3cos θ

Forgetting the denominator in tangent identities

Tangent double-angle and triple-angle formulas are rational expressions.

Ignoring undefined values

Tangent identities require their denominators to be nonzero.

Forgetting the ± sign in half-angle identities

The sign depends on the quadrant of the half angle.

Using the quadrant of θ instead of θ/2

Always determine the quadrant of the angle being evaluated.

Rounding exact values too early

Keep fractions and radicals until the final answer.

Dividing by trigonometric factors when solving equations

This can eliminate valid solutions.

Failing to adjust intervals

For sin 3x, the angle 3x covers three times the interval of x.

Confusing frequency and amplitude

Increasing the angle multiplier changes the period, not the amplitude of the standard sine or cosine function.

Key Terms

Multiple-Angle Identity: A trigonometric identity involving an angle multiplied by a number.

Double Angle: An angle twice as large as another angle.

Triple Angle: An angle three times as large as another angle.

Half Angle: An angle equal to half another angle.

Double-Angle Identity: An identity expressing a trigonometric function of 2θ using functions of θ.

Triple-Angle Identity: An identity expressing a trigonometric function of 3θ using functions of θ.

Half-Angle Identity: An identity expressing a trigonometric function of θ/2 using functions of θ.

Sine Triple-Angle Identity: The relationship sin 3θ = 3sin θ − 4sin³ θ.

Cosine Triple-Angle Identity: The relationship cos 3θ = 4cos³ θ − 3cos θ.

Tangent Triple-Angle Identity: The relationship tan 3θ = (3tan θ − tan³ θ)/(1 − 3tan² θ), where defined.

Power-Reduction Identity: An identity that rewrites squared trigonometric functions using double angles.

Compound Angle Identity: An identity involving the sum or difference of two angles.

Pythagorean Identity: A trigonometric relationship derived from sin² θ + cos² θ = 1.

Equivalent Expressions: Expressions that produce the same values over their common domain.

Identity Proof: A logical demonstration that two mathematical expressions are equivalent.

Verification: Checking that a mathematical statement or proposed solution is valid.

Exact Value: A mathematical value expressed without rounding.

Reference Angle: The acute angle between an angle's terminal side and the x-axis.

Quadrant: One of the four regions of the Cartesian coordinate plane.

Unit Circle: A circle of radius 1 centred at the origin.

Principal Square Root: The nonnegative square root of a nonnegative number.

Factorisation: Rewriting an expression as a product of factors.

Algebraic Substitution: Replacing an expression with an equivalent expression.

Domain Restriction: A condition excluding values for which an expression is undefined.

Period: The horizontal length of one complete cycle of a periodic function.

Frequency: The number of cycles completed per unit time.

Angular Frequency: The rate of angular change, commonly represented by ω.

Amplitude: The maximum displacement of a sinusoidal function from its midline.

Harmonic: A sinusoidal component whose frequency is an integer multiple of a fundamental frequency.

Sinusoidal Function: A function represented using sine or cosine.

Simple Harmonic Motion: Oscillatory motion that can be modelled using sine or cosine functions.

Antiderivative: A function whose derivative equals a given function.

Integration: A mathematical process used to calculate accumulated quantities and areas.

Check Your Understanding

1. Explain the difference between sin 3θ and sin³ θ.

2. Derive the sine double-angle identity from the sine addition identity.

3. Write the three equivalent forms of the cosine double-angle identity.

4. Derive the tangent double-angle identity.

5. Use a double-angle identity to calculate sin 120° exactly.

6. Derive the sine triple-angle identity.

7. Derive the cosine triple-angle identity.

8. Given sin θ = 3/5 and θ is acute, determine sin 3θ.

9. Given cos θ = 4/5 and θ is acute, determine cos 3θ.

10. Determine the exact value of sin 22.5° using a half-angle identity.

11. Simplify:

3sin θ − 4sin³ θ

12. Simplify:

4cos³ θ − 3cos θ

13. Verify:

sin 3θ − sin θ = 2sin θ cos 2θ

14. Solve:

sin 3x = sin x

for 0° ≤ x < 360°.

15. Solve:

cos 3x = cos x

for 0° ≤ x < 360°.

Key Takeaways

• Multiple-angle identities connect trigonometric functions of double, triple, and half angles with functions of the original angle.

• Double-angle identities are derived from compound angle identities by setting both angles equal.

• The sine double-angle identity is sin 2θ = 2sin θ cos θ.

• The cosine double-angle identity has three equivalent forms.

• The tangent double-angle identity is tan 2θ = 2tan θ/(1 − tan² θ), where defined.

• Triple-angle identities can be derived by writing 3θ = 2θ + θ.

• The sine triple-angle identity is sin 3θ = 3sin θ − 4sin³ θ.

• The cosine triple-angle identity is cos 3θ = 4cos³ θ − 3cos θ.

• The tangent triple-angle identity expresses tan 3θ using tan θ.

• Half-angle identities are derived by rearranging cosine double-angle identities.

• Half-angle identities are useful for calculating exact values such as 15° and 22.5°.

• The sign in a half-angle identity depends on the quadrant containing θ/2.

• Multiple-angle identities can be rearranged into equivalent forms using Pythagorean identities.

• Power-reduction identities express sin² θ and cos² θ using cos 2θ.

• Recognising multiple-angle patterns helps simplify complicated expressions.

• Double-angle and triple-angle identities are useful for solving trigonometric equations.

• Equations involving multiple angles may have more solutions within a given interval.

• Factoring helps preserve all possible solutions when solving trigonometric equations.

• Tangent identities require careful attention to undefined values.

• Multiple-angle identities are important in geometry, wave motion, oscillations, and signal analysis.

• Power-reduction identities are especially useful in calculus and integration.

• Understanding how multiple-angle identities are derived provides a strong foundation for advanced trigonometry, mathematical modelling, physics, engineering, and calculus.