Introduction to Double Angle Formulas
The double angle formulas express trigonometric functions of 2θ in terms of functions of θ. They are not new, independent identities; rather, they are special cases of the angle addition formulas (sum formulas) where both angles happen to be the same.
These formulas are indispensable in advanced trigonometry, precalculus, and calculus. They appear in integration techniques, solving equations, simplifying expressions, and modeling periodic phenomena. Understanding where they come from (the sum formulas) makes them far easier to remember and apply.
Derivation from the Sum Formulas
Recall the angle addition (sum) formulas:
cos(A + B) = cos A cos B − sin A sin B
tan(A + B) = (tan A + tan B) / (1 − tan A tan B)
Now set A = B = θ in each formula. This substitution is the entire derivation process.
The Double Angle Formula for Sine
Starting with sin(A + B) = sin A cos B + cos A sin B and substituting A = B = θ:
This is the sine double angle formula. It has exactly one form, and it is beautifully simple: doubling the angle doubles the product of sine and cosine.
This formula is used frequently when you need to express a product sin θ cos θ as a single trig function, or when you see sin(2θ) and want to break it into factors involving θ.
The Double Angle Formulas for Cosine
Starting with cos(A + B) = cos A cos B − sin A sin B and substituting A = B = θ:
This is the primary form. Using the Pythagorean identity sin²θ + cos²θ = 1, we can substitute to produce two additional equivalent forms:
Form 2: Replace sin²θ with 1 − cos²θ
Form 3: Replace cos²θ with 1 − sin²θ
All three forms are equally valid. The form you choose depends on which functions appear in your problem:
- cos²θ − sin²θ: Use when both sine and cosine are present.
- 2cos²θ − 1: Use when only cosine appears or you want to eliminate sine.
- 1 − 2sin²θ: Use when only sine appears or you want to eliminate cosine.
Key Takeaway
The cosine double angle formula has three equivalent forms. Choose the one that matches the functions already present in your problem. All three are derived from a single formula plus the Pythagorean identity.
The Double Angle Formula for Tangent
From the tangent sum formula with A = B = θ:
Note that this formula is undefined when tan²θ = 1, which occurs at θ = 45° + n · 90°. At those values, tan(2θ) itself is undefined because 2θ is an odd multiple of 90°.
Half Angle Formulas
The half angle formulas are derived by rearranging the double angle formulas for cosine. They express trig functions of θ/2 in terms of functions of θ.
Deriving sin(θ/2)
Start with cos(2α) = 1 − 2sin²α and let α = θ/2:
Solving for sin(θ/2):
Deriving cos(θ/2)
Start with cos(2α) = 2cos²α − 1 and let α = θ/2:
Deriving tan(θ/2)
Dividing the sine half angle by the cosine half angle, or using algebraic manipulation, we get two useful forms:
tan(θ/2) = sin θ / (1 + cos θ) = (1 − cos θ) / sin θ
The last two forms (without the ± sign) are often more convenient because they automatically give the correct sign based on the signs of sine and cosine.
Key Takeaway
The ± in the half angle formulas is determined by the quadrant where θ/2 lies, not where θ lies. Always determine the quadrant of the half angle before choosing the sign.
Power-Reducing Identities
The power-reducing identities (also called power-reduction formulas) express sin²θ and cos²θ in terms of the first power of cosine of a double angle. They are direct rearrangements of the double angle formulas for cosine:
cos²θ = (1 + cos 2θ) / 2
tan²θ = (1 − cos 2θ) / (1 + cos 2θ)
These identities are essential in calculus for integrating even powers of sine and cosine. Without them, integrals like ∫ sin²(x) dx and ∫ cos&sup4;(x) dx would be much more difficult to evaluate.
Product-to-Sum Formulas
The product-to-sum formulas convert products of sines and cosines into sums, which are often easier to integrate or simplify:
cos A cos B = (1/2)[cos(A−B) + cos(A+B)]
sin A sin B = (1/2)[cos(A−B) − cos(A+B)]
These are derived by adding or subtracting the angle addition and subtraction formulas. They are particularly useful in signal processing, Fourier analysis, and integral calculus where products of trig functions appear frequently.
Applications in Calculus
Double angle and power-reducing formulas find their most frequent application in integral calculus. Here is why they matter:
- Integrating sin²(x): The integral ∫ sin²(x) dx cannot be computed directly. Using the power-reducing identity sin²(x) = (1 − cos 2x)/2, the integral becomes ∫ (1 − cos 2x)/2 dx = x/2 − sin(2x)/4 + C.
- Integrating cos²(x): Similarly, ∫ cos²(x) dx = ∫ (1 + cos 2x)/2 dx = x/2 + sin(2x)/4 + C.
- Higher even powers: For cos&sup4;(x), apply the power-reducing identity twice to reduce all powers to first-degree cosines.
- Trigonometric substitution: The double angle formula sin(2θ) = 2 sin θ cos θ often appears when back-substituting after a trig substitution in integration.
Worked Examples
Example 1: Find sin(2θ) Given sin θ
Given: sin θ = 3/5 and θ is in Quadrant I. Find sin(2θ).
Step 1: Find cos θ using the Pythagorean identity.
cos²θ = 1 − sin²θ = 1 − 9/25 = 16/25
cos θ = 4/5 (positive because θ is in QI)
Step 2: Apply the double angle formula.
sin(2θ) = 2 sin θ cos θ = 2(3/5)(4/5) = 24/25
Example 2: Simplify cos²(x) for Integration
Problem: Evaluate ∫ cos²(x) dx.
Step 1: Apply the power-reducing identity.
cos²(x) = (1 + cos 2x) / 2
Step 2: Integrate term by term.
∫ (1 + cos 2x)/2 dx = (1/2) ∫ 1 dx + (1/2) ∫ cos 2x dx
Step 3: Evaluate each integral.
= x/2 + (1/2) · sin(2x)/2 + C = x/2 + sin(2x)/4 + C
Example 3: Evaluate sin(15°) Using Half Angle
Goal: Find the exact value of sin(15°) using the half angle formula.
Step 1: Recognize that 15° = 30°/2, so θ = 30° in the half angle formula.
sin(15°) = sin(30°/2) = √[(1 − cos 30°) / 2]
Step 2: The sign is positive because 15° is in Quadrant I.
Step 3: Substitute cos 30° = √3/2.
sin(15°) = √[(1 − √3/2) / 2] = √[(2 − √3) / 4]
Step 4: Simplify.
sin(15°) = √(2 − √3) / 2 ≈ 0.2588
Example 4: Find cos(2θ) Given tan θ
Given: tan θ = 5/12 and θ is in Quadrant III. Find cos(2θ).
Step 1: Find sin θ and cos θ. In QIII, both are negative.
With tan θ = 5/12, the reference triangle has opposite = 5, adjacent = 12, hypotenuse = 13.
sin θ = −5/13, cos θ = −12/13
Step 2: Apply cos(2θ) = cos²θ − sin²θ.
cos(2θ) = (−12/13)² − (−5/13)² = 144/169 − 25/169 = 119/169
Practice Problems
Practice 1
If cos θ = −7/25 and θ is in Quadrant II, find sin(2θ), cos(2θ), and tan(2θ).
Solution:
In QII, sin θ is positive. sin θ = √(1 − 49/625) = √(576/625) = 24/25
sin(2θ) = 2(24/25)(−7/25) = −336/625
cos(2θ) = (−7/25)² − (24/25)² = 49/625 − 576/625 = −527/625
tan(2θ) = sin(2θ)/cos(2θ) = (−336/625) / (−527/625) = 336/527
Practice 2
Find the exact value of cos(22.5°) using the half angle formula.
Solution:
22.5° = 45°/2, so θ = 45°.
cos(22.5°) = √[(1 + cos 45°)/2] (positive since 22.5° is in QI)
= √[(1 + √2/2)/2] = √[(2 + √2)/4]
= √(2 + √2) / 2 ≈ 0.9239
Practice 3
Solve the equation sin(2x) = cos(x) for 0 ≤ x < 2π.
Solution:
Replace sin(2x) with 2 sin x cos x:
2 sin x cos x = cos x
2 sin x cos x − cos x = 0
cos x (2 sin x − 1) = 0
cos x = 0 → x = π/2, 3π/2
sin x = 1/2 → x = π/6, 5π/6
Solutions: x = π/6, π/2, 5π/6, 3π/2
Practice 4
Use a power-reducing identity to rewrite sin²(x) cos²(x) as a single trig function without exponents.
Solution:
Method: Use the identity sin(2x) = 2 sin x cos x, so sin x cos x = sin(2x)/2.
sin²(x) cos²(x) = [sin x cos x]² = [sin(2x)/2]² = sin²(2x)/4
Now apply power-reducing to sin²(2x):
= (1/4) · (1 − cos 4x)/2 = (1 − cos 4x) / 8