Double-Angle and Half-Angle Identities

Double-angle and half-angle identities are powerful tools in trigonometry that allow us to rewrite expressions involving angles that are twice or half the size of another angle.

For example, if we know sin 30° and cos 30°, we can calculate sin 60° using a double-angle identity. Similarly, we can determine the exact value of sin 15° using a half-angle identity.

These identities build directly on the sum and difference identities and are especially useful for:

• Calculating exact trigonometric values.

• Simplifying complicated trigonometric expressions.

• Verifying and deriving equivalent identities.

• Solving trigonometric equations.

• Analysing waves, oscillations, and periodic motion.

• Performing advanced calculations in calculus and physics.

Understanding Double Angles and Half Angles

A double angle is an angle that is twice another angle.

Examples include:

• 2 × 30° = 60°

• 2 × 45° = 90°

• 2 × 60° = 120°

• 2 × π/6 = π/3

A half angle is an angle that is half another angle.

Examples include:

• 60°/2 = 30°

• 90°/2 = 45°

• 30°/2 = 15°

• (π/3)/2 = π/6

The notation sin 2θ means the sine of twice the angle θ.

It is important to distinguish between:

sin 2θ = sin(2θ)

and:

sin² θ = (sin θ)²

These expressions are generally not equal.

For example, when θ = 30°:

sin 2θ = sin 60° = √3/2

sin² θ = (1/2)² = 1/4

Understanding this distinction is essential when applying trigonometric identities.

Deriving the Sine Double-Angle Identity

Recall the sine addition identity:

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

To derive the double-angle identity, let:

A = θ

B = θ

Then:

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

Since:

θ + θ = 2θ

we obtain:

sin 2θ = sin θ cos θ + sin θ cos θ

Combine like terms:

sin 2θ = 2 sin θ cos θ

This is the sine double-angle identity.

Worked Example: Finding sin 60°

Let:

θ = 30°

Use:

sin 2θ = 2 sin θ cos θ

Therefore:

sin 60° = 2 sin 30° cos 30°

Substitute:

= 2(1/2)(√3/2)

Simplify:

sin 60° = √3/2

This agrees with the known exact value.

Worked Example: Finding sin 120°

Let:

θ = 60°

sin 120° = 2 sin 60° cos 60°

= 2(√3/2)(1/2)

Therefore:

sin 120° = √3/2

The result is positive, as expected in Quadrant II.

Deriving the Cosine Double-Angle Identity

Recall the cosine addition identity:

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

Let:

A = θ

B = θ

Then:

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

Therefore:

cos 2θ = cos² θ − sin² θ

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

However, cosine has two additional useful forms.

Deriving the Second Form

Begin with:

cos 2θ = cos² θ − sin² θ

Use the Pythagorean identity:

sin² θ + cos² θ = 1

Therefore:

sin² θ = 1 − cos² θ

Substitute:

cos 2θ = cos² θ − (1 − cos² θ)

Expand:

= cos² θ − 1 + cos² θ

Combine:

cos 2θ = 2cos² θ − 1

Deriving the Third Form

Again, begin 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

The Three Forms

All three expressions are equivalent:

cos 2θ = cos² θ − sin² θ

cos 2θ = 2cos² θ − 1

cos 2θ = 1 − 2sin² θ

Each form is useful in different situations.

Applying Cosine Double-Angle Identities

Worked Example: Finding cos 60°

Let:

θ = 30°

Use:

cos 2θ = 2cos² θ − 1

Then:

cos 60° = 2cos² 30° − 1

Substitute:

= 2(√3/2)² − 1

= 2(3/4) − 1

= 3/2 − 1

cos 60° = 1/2

Worked Example: Finding cos 120°

Let:

θ = 60°

Use:

cos 2θ = 1 − 2sin² θ

Then:

cos 120° = 1 − 2sin² 60°

= 1 − 2(√3/2)²

= 1 − 2(3/4)

= 1 − 3/2

cos 120° = −1/2

The result is negative, consistent with Quadrant II.

Deriving the Tangent Double-Angle Identity

Recall the tangent addition identity:

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

Let:

A = θ

B = θ

Then:

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

Combine:

tan 2θ = 2tan θ/(1 − tan² θ)

This is the tangent double-angle identity.

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

Worked Example: Finding tan 60°

Let:

θ = 30°

tan 60° = 2tan 30°/(1 − tan² 30°)

Substitute:

= 2(√3/3)/(1 − (√3/3)²)

= (2√3/3)/(1 − 1/3)

= (2√3/3)/(2/3)

Therefore:

tan 60° = √3

Worked Example: Using a Known Tangent Value

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)

tan 2θ = 3/4

Summary of Double-Angle Identities

These identities can be used in both directions.

For example:

2sin θ cos θ = sin 2θ

and:

1 − 2sin² θ = cos 2θ

Recognising these patterns is particularly useful when simplifying expressions.

Finding Double-Angle Values from Given Ratios

Double-angle identities can be applied even when the angle itself is unknown.

Worked Example

Suppose:

sin θ = 3/5

and θ is acute.

Find:

• cos θ

• sin 2θ

• cos 2θ

• tan 2θ

Find cos θ

Use:

sin² θ + cos² θ = 1

Substitute:

(3/5)² + cos² θ = 1

cos² θ = 1 − 9/25

cos² θ = 16/25

Since θ is acute:

cos θ = 4/5

Find sin 2θ

sin 2θ = 2sin θ cos θ

= 2(3/5)(4/5)

sin 2θ = 24/25

Find cos 2θ

cos 2θ = cos² θ − sin² θ

= (4/5)² − (3/5)²

= 16/25 − 9/25

cos 2θ = 7/25

Find tan 2θ

tan 2θ = sin 2θ/cos 2θ

= (24/25)/(7/25)

tan 2θ = 24/7

All three values were calculated without finding θ.

The Importance of Quadrants

Double-angle calculations may require careful consideration of signs.

For example, suppose:

cos θ = −3/5

and θ lies in Quadrant II.

Find sin 2θ.

First, determine sine.

sin² θ = 1 − cos² θ

= 1 − 9/25

= 16/25

Since sine is positive in Quadrant II:

sin θ = 4/5

Now:

sin 2θ = 2sin θ cos θ

= 2(4/5)(−3/5)

sin 2θ = −24/25

This makes sense because doubling an angle in Quadrant II produces an angle between 180° and 360°, where sine is negative.

Deriving the Half-Angle Identities

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

Start with:

cos 2θ = 1 − 2sin² θ

Rearrange:

2sin² θ = 1 − cos 2θ

Divide by 2:

sin² θ = (1 − cos 2θ)/2

Now replace θ with θ/2:

sin²(θ/2) = (1 − cos θ)/2

Take the square root:

sin(θ/2) = ±√[(1 − cos θ)/2]

This is the sine half-angle identity.

Why Is There a ± Sign?

Taking the square root of a squared expression produces positive and negative possibilities.

The correct sign depends on the quadrant containing θ/2.

For example:

sin 15° is positive because 15° lies in Quadrant I.

However:

sin 195° is negative because 195° lies in Quadrant III.

The half-angle formula must therefore account for the sign.

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]

The correct sign depends on the quadrant containing θ/2.

Deriving the Tangent Half-Angle Identity

Tangent can be written as:

tan(θ/2) = sin(θ/2)/cos(θ/2)

Using the half-angle identities gives:

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

where the expression is defined.

However, two additional forms are often easier to use.

From the double-angle identities:

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)]

= sin(θ/2)/cos(θ/2)

Therefore:

tan(θ/2) = sin θ/(1 + cos θ)

Similarly:

tan(θ/2) = (1 − cos θ)/sin θ

These two forms are equivalent wherever their respective expressions are defined.

Summary of Half-Angle Identities

The ± sign must be selected using the quadrant of θ/2.

Applying Half-Angle Identities to Exact Values

Half-angle identities are particularly useful for finding exact values of angles such as 15°, 22.5°, and 75°.

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 the fraction:

= √[(2 − √3)/4]

Therefore:

sin 15° = √(2 − √3)/2

This is equivalent to the sum-and-difference result:

sin 15° = (√6 − √2)/4

Both are exact values.

Worked Example: Finding cos 15°

Use:

cos(θ/2) = ±√[(1 + cos θ)/2]

Let:

θ = 30°

Then:

cos 15° = √[(1 + √3/2)/2]

Simplify:

= √[(2 + √3)/4]

Therefore:

cos 15° = √(2 + √3)/2

An equivalent expression is:

cos 15° = (√6 + √2)/4

Finding Exact Values of 22.5°

The angle 22.5° is half of 45°.

This makes it suitable for half-angle identities.

Worked Example: Finding sin 22.5°

Use:

sin(θ/2) = ±√[(1 − cos θ)/2]

Let:

θ = 45°

Since 22.5° lies in Quadrant I, choose the positive square root.

sin 22.5° = √[(1 − cos 45°)/2]

Substitute:

= √[(1 − √2/2)/2]

Simplify:

= √[(2 − √2)/4]

Therefore:

sin 22.5° = √(2 − √2)/2

Worked Example: Finding cos 22.5°

Use:

cos(θ/2) = ±√[(1 + cos θ)/2]

Then:

cos 22.5° = √[(1 + √2/2)/2]

Simplify:

cos 22.5° = √(2 + √2)/2

Finding Half-Angle Values from Known Ratios

Suppose:

cos θ = −3/5

and:

90° < θ < 180°

Find sin(θ/2), cos(θ/2), and tan(θ/2).

Determine the Quadrant

Since θ lies in Quadrant II:

45° < θ/2 < 90°

Therefore, θ/2 lies in Quadrant I.

All three trigonometric functions are positive.

Find sin(θ/2)

Use:

sin(θ/2) = √[(1 − cos θ)/2]

Substitute:

= √[(1 − (−3/5))/2]

= √[(1 + 3/5)/2]

= √[(8/5)/2]

= √(4/5)

Therefore:

sin(θ/2) = 2√5/5

Find cos(θ/2)

Use:

cos(θ/2) = √[(1 + cos θ)/2]

Substitute:

= √[(1 − 3/5)/2]

= √[(2/5)/2]

= √(1/5)

Therefore:

cos(θ/2) = √5/5

Find tan(θ/2)

Use:

tan(θ/2) = sin(θ/2)/cos(θ/2)

= (2√5/5)/(√5/5)

Therefore:

tan(θ/2) = 2

Power-Reduction Identities

The cosine double-angle identities can be rearranged to rewrite squared trigonometric functions.

These are called power-reduction identities.

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 identities replace squared sine or cosine functions with expressions involving a double angle.

Why Are They Useful?

Power-reduction identities are especially useful in calculus because they allow expressions such as sin² x and cos² x to be rewritten in forms that are easier to integrate.

Worked Example

Simplify:

2sin² x − 1

Use:

cos 2x = 1 − 2sin² x

Multiply both sides by −1:

−cos 2x = −1 + 2sin² x

Therefore:

2sin² x − 1 = −cos 2x

Worked Example

Rewrite:

cos² x

using a double angle.

Apply:

cos² x = (1 + cos 2x)/2

This form is often more convenient in integration problems.

Simplifying Expressions Using Double-Angle Identities

Worked Example

Simplify:

4sin x cos x

Use:

sin 2x = 2sin x cos x

Therefore:

4sin x cos x = 2(2sin x cos x)

= 2sin 2x

Worked Example

Simplify:

cos² x − sin² x

Use the cosine double-angle identity.

= cos 2x

Worked Example

Simplify:

1 − 2cos² x

Since:

cos 2x = 2cos² x − 1

Multiply by −1:

1 − 2cos² x = −cos 2x

Worked Example

Simplify:

2sin² x + cos 2x

Use:

cos 2x = 1 − 2sin² x

Substitute:

= 2sin² x + 1 − 2sin² x

Therefore:

= 1

Verifying Equivalent Identities

Double-angle and half-angle identities can be used to verify more complicated relationships.

Worked Example

Verify:

sin 2x/(1 + cos 2x) = tan x

Start with the left-hand side:

sin 2x/(1 + cos 2x)

Use:

sin 2x = 2sin x cos x

and:

cos 2x = 2cos² x − 1

Substitute:

= 2sin x cos x/[1 + 2cos² x − 1]

Simplify:

= 2sin x cos x/(2cos² x)

Cancel common factors:

= sin x/cos x

Therefore:

= tan x

The identity is verified wherever cos x ≠ 0.

Worked Example

Verify:

(1 − cos 2x)/sin 2x = tan x

Start with:

(1 − cos 2x)/sin 2x

Use:

1 − cos 2x = 2sin² x

and:

sin 2x = 2sin x cos x

Substitute:

= 2sin² x/(2sin x cos x)

Simplify:

= sin x/cos x

Therefore:

= tan x

This identity is verified where sin x ≠ 0 and cos x ≠ 0, which is the domain of the original left-hand side.

Deriving Additional Equivalent Identities

Trigonometric identities can often be rearranged to produce new relationships.

Example: Expressing sin² θ in Terms of cos 2θ

Start with:

cos 2θ = 1 − 2sin² θ

Rearrange:

2sin² θ = 1 − cos 2θ

Therefore:

sin² θ = (1 − cos 2θ)/2

Example: Expressing cos² θ in Terms of cos 2θ

Start with:

cos 2θ = 2cos² θ − 1

Rearrange:

2cos² θ = 1 + cos 2θ

Therefore:

cos² θ = (1 + cos 2θ)/2

Example: Deriving an Identity for tan 2θ Using Sine and Cosine

Start with:

tan 2θ = sin 2θ/cos 2θ

Substitute the double-angle identities:

= 2sin θ cos θ/(cos² θ − sin² θ)

Therefore:

tan 2θ = 2sin θ cos θ/(cos² θ − sin² θ)

This form can be useful when sine and cosine are known directly.

Solving Trigonometric Equations Using Double Angles

Double-angle identities can simplify equations involving multiple angles.

Worked Example

Solve:

sin 2x = 0

for:

0° ≤ x < 360°

We know:

sin θ = 0

when:

θ = 180°n

where n is any integer.

Therefore:

2x = 180°n

Divide by 2:

x = 90°n

Within the required interval:

x = 0°, 90°, 180°, 270°

Worked Example

Solve:

cos 2x = 1/2

for:

0° ≤ x < 360°

Since x ranges from 0° to 360°, 2x ranges from 0° to 720°.

Cosine equals 1/2 at:

2x = 60°, 300°, 420°, 660°

Divide each by 2:

x = 30°, 150°, 210°, 330°

Solving Equations by Rewriting Double Angles

Worked Example

Solve:

2sin x cos x = sin x

for:

0° ≤ x < 360°

Recognise:

2sin x cos x = sin 2x

However, factoring is the simplest method.

Move everything to one side:

2sin x cos x − sin x = 0

Factor:

sin x(2cos x − 1) = 0

Therefore:

sin x = 0

or:

2cos x − 1 = 0

First Case

sin x = 0

Therefore:

x = 0°, 180°

Second Case

2cos x − 1 = 0

cos x = 1/2

Therefore:

x = 60°, 300°

Final Answer

x = 0°, 60°, 180°, 300°

Important Principle

When an equation contains products of trigonometric functions, factoring may be more effective than dividing.

Dividing by sin x would incorrectly eliminate solutions where sin x = 0.

Solving Equations Using Half-Angle Relationships

Worked Example

Solve:

sin² x = 1/4

for:

0° ≤ x < 360°

Take the square root:

sin x = ±1/2

Sine equals 1/2 at:

30°, 150°

Sine equals −1/2 at:

210°, 330°

Therefore:

x = 30°, 150°, 210°, 330°

Alternatively, the equation can be rewritten using:

sin² x = (1 − cos 2x)/2

Substitute:

(1 − cos 2x)/2 = 1/4

Multiply by 2:

1 − cos 2x = 1/2

Therefore:

cos 2x = 1/2

This produces the same solutions.

Double Angles and Graph Transformations

Double-angle identities also connect to the graphs of trigonometric functions.

Consider:

y = sin x

and:

y = sin 2x

The standard sine function has period:

2π

The function sin 2x has period:

2π/2 = π

Therefore, y = sin 2x completes twice as many cycles over the same horizontal interval.

Important Distinction

The identity:

sin 2x = 2sin x cos x

does not mean that doubling the angle doubles the amplitude.

The function y = sin 2x still has amplitude 1.

The expression 2sin x cos x always remains between −1 and 1.

Applications in Physics: Waves and Oscillations

Double-angle identities are useful when studying periodic motion.

For example, a wave may be represented by:

y(t) = A sin(ωt)

A wave with twice the angular frequency is:

y(t) = A sin(2ωt)

Using the double-angle identity:

sin(2ωt) = 2sin(ωt)cos(ωt)

Therefore:

y(t) = 2A sin(ωt)cos(ωt)

This relationship can be useful when analysing combinations of oscillatory signals.

Frequency and Period

For:

y(t) = A sin(ωt)

the period is:

P = 2π/ω

For:

y(t) = A sin(2ωt)

the period becomes:

P = 2π/(2ω)

P = π/ω

Doubling angular frequency halves the period.

Applications in Calculus

Power-reduction identities are especially useful when integrating squared trigonometric functions.

For example:

∫ sin² x dx

This is difficult to integrate directly using basic antiderivatives.

Use:

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 formulas appear in calculations involving wave energy, oscillations, and periodic systems.

Applications in Geometry

Double-angle and half-angle identities can help determine exact angles and lengths in geometric problems.

Worked Example: Finding an Exact Half Angle

Suppose a triangle contains an angle of 45°.

We want to determine the exact sine of half that angle.

Half the angle:

45°/2 = 22.5°

Use:

sin(θ/2) = √[(1 − cos θ)/2]

Substitute:

sin 22.5° = √[(1 − √2/2)/2]

Therefore:

sin 22.5° = √(2 − √2)/2

This value can be used to calculate lengths in triangles containing 22.5° angles.

Example: Finding a Side Length

Suppose a right triangle has:

• Hypotenuse = 10 cm

• Acute angle = 22.5°

The opposite side is:

a = 10 sin 22.5°

Substitute the exact value:

a = 10[√(2 − √2)/2]

Therefore:

a = 5√(2 − √2) cm

Approximately:

a ≈ 3.83 cm

Choosing the Best Identity

Different expressions suggest different identities.

Recognising these patterns can make complicated calculations much shorter.

A Strategy for Solving Double-Angle and Half-Angle Problems

When solving problems involving multiple angles:

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

2. Determine which trigonometric function is involved.

3. Select the most useful identity.

4. Substitute known values or equivalent expressions.

5. Check the quadrant before selecting signs in half-angle formulas.

6. Simplify algebraically.

7. Factor expressions when solving equations.

8. Avoid dividing by expressions that could equal zero.

9. Check the required angle interval.

10. Verify results using the original equation when possible.

For exact-value problems, keep radicals and fractions rather than using decimal approximations.

Common Mistakes

Confusing sin 2θ with sin² θ

sin 2θ means sin(2θ), while sin² θ means (sin θ)².

Assuming sin 2θ = 2sin θ

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.

Forgetting the denominator in tangent double-angle calculations

tan 2θ = 2tan θ/(1 − tan² θ)

Ignoring undefined values

The tangent double-angle quotient formula cannot be used when its denominator is zero or tan θ is undefined.

Forgetting the ± sign in half-angle identities

The correct sign depends on the quadrant of the half angle.

Using the quadrant of θ instead of θ/2

Always determine the quadrant of the actual angle being evaluated.

Assuming double angles always remain in the same quadrant

Doubling an angle can move it into a different quadrant.

Rounding exact values too early

Keep radicals and fractions until the final answer.

Dividing by trigonometric factors when solving equations

This can eliminate valid solutions.

Forgetting the full interval when solving equations

If the equation contains 2x, the corresponding angle interval must also be doubled.

Confusing frequency and amplitude

Doubling the angle changes the period, not the amplitude of a standard sine or cosine function.

Key Terms

Double Angle: An angle that is twice another angle.

Half Angle: An angle that is half another angle.

Double-Angle Identity: A trigonometric identity expressing a function of 2θ in terms of functions of θ.

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

Sine Double-Angle Identity: The relationship sin 2θ = 2sin θ cos θ.

Cosine Double-Angle Identity: Any of the equivalent formulas expressing cos 2θ using sine and cosine of θ.

Tangent Double-Angle Identity: The relationship tan 2θ = 2tan θ/(1 − tan² θ), where defined.

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

Compound Angle: An angle formed by adding or subtracting two angles.

Trigonometric Identity: An equation involving trigonometric functions that is true wherever both sides are defined.

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

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

Exact Value: A mathematical value expressed without rounding.

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

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

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

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

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

Frequency: The number of complete cycles occurring 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.

Sinusoidal Function: A function represented using sine or cosine.

Trigonometric Equation: An equation involving trigonometric functions that is true for particular values of a variable.

Verification: Demonstrating that a proposed identity is mathematically valid.

Factorisation: Rewriting an expression as a product of factors.

Antiderivative: A function whose derivative equals a given function.

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

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

Check Your Understanding

1. Explain the difference between sin 2θ and sin² θ.

2. State the sine double-angle identity.

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

4. State the tangent double-angle identity.

5. Use a double-angle identity to calculate sin 60° exactly.

6. Use a double-angle identity to calculate cos 120° exactly.

7. Given sin θ = 5/13 and θ is acute, determine sin 2θ and cos 2θ.

8. Given tan θ = 2/3, determine tan 2θ.

9. Derive the sine half-angle identity from the cosine double-angle identity.

10. Determine the exact value of sin 22.5°.

11. Determine the exact value of cos 22.5°.

12. Simplify:

4sin x cos x

13. Verify:

(1 − cos 2x)/sin 2x = tan x

State the domain restrictions.

14. Solve:

cos 2x = −1/2

for 0° ≤ x < 360°.

15. A right triangle has a hypotenuse of 16 cm and an acute angle of 22.5°. Use a half-angle identity to determine the exact length of the side opposite the angle.

Key Takeaways

• Double-angle and half-angle identities allow trigonometric expressions involving multiple angles to be rewritten and simplified.

• Double-angle identities are derived from the sum 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.

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

• The sine half-angle identity is sin(θ/2) = ±√[(1 − cos θ)/2].

• The cosine half-angle identity is cos(θ/2) = ±√[(1 + cos θ)/2].

• Tangent half-angle identities can be expressed using sine and cosine of the original angle.

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

• Double-angle identities can calculate trigonometric values without finding the original angle.

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

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

• Trigonometric expressions can often be simplified by recognising double-angle patterns.

• Double-angle identities can be used to verify and derive equivalent identities.

• Trigonometric equations involving 2x may have more solutions over a given interval than equations involving x.

• Factoring helps preserve all solutions when solving trigonometric equations.

• Doubling the angle in sine or cosine changes the period but does not change the standard amplitude.

• Double-angle and half-angle identities are important in geometry, waves, oscillations, and physics.

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

• Mastering these identities develops the skills needed for advanced trigonometric problem-solving, mathematical modelling, and calculus.