7.3Applying Double-Angle and Half-Angle Formulas
Apply sin(2θ) = 2sinθcosθ, the three forms of cos(2θ), half-angle formulas, and power-reduction identities to simplify expressions and find exact values.
Double-angle and power-reduction formulas are used constantly in calculus to integrate powers of sine and cosine. They are among the most practically important identities in all of trigonometry.
How do the double-angle and half-angle formulas extend the sum/difference identities, and why are they so useful in calculus?
The double-angle formulas are special cases of the sum formulas with α = β = θ. The half-angle formulas are derived by solving the cos(2θ) identities for sinθ or cosθ.
Power-Reduction Formulas
sin²θ = (1 − cos2θ)/2
cos²θ = (1 + cos2θ)/2
tan²θ = (1 − cos2θ)/(1 + cos2θ)
Derivation of sin²θ
From cos(2θ) = 1 − 2sin²θ:
2sin²θ = 1 − cos(2θ)
sin²θ = (1 − cos2θ)/2
Given sinθ = 3/5 and θ is in QI, find sin(2θ) and cos(2θ).
Find cosθ: cos²θ = 1 − (3/5)² = 16/25 → cosθ = 4/5 (QI).
sin(2θ) = 2sinθcosθ = 2(3/5)(4/5) = 24/25.
cos(2θ) = cos²θ − sin²θ = 16/25 − 9/25 = 7/25.
Find the exact value of sin(π/8).
Use the half-angle formula: sin(θ/2) = ±√((1 − cosθ)/2) with θ = π/4.
cos(π/4) = √2/2.
sin(π/8) = √((1 − √2/2)/2) = √((2 − √2)/4) = √(2 − √2)/2.
Since π/8 is in QI, the sign is positive.
Simplify: 2sin(3x)cos(3x)
Recognize the double-angle pattern: 2sinθcosθ = sin(2θ) with θ = 3x.
2sin(3x)cos(3x) = sin(2 · 3x) = sin(6x).
Use a power-reduction formula to rewrite sin²x cos²x as a function of cos(4x).
sin²x cos²x = (sinx cosx)² = (sin(2x)/2)² = sin²(2x)/4.
Apply power-reduction to sin²(2x): sin²(2x) = (1 − cos(4x))/2.
sin²x cos²x = (1 − cos(4x))/8.
Given cosθ = −1/3 and π {'<'} θ {'<'} 3π/2 (QIII), find cos(θ/2).
Use cos(θ/2) = ±√((1 + cosθ)/2).
cos(θ/2) = ±√((1 + (−1/3))/2) = ±√((2/3)/2) = ±√(1/3) = ±1/√3 = ±√3/3.
Determine the sign: θ is in QIII, so π < θ < 3π/2, meaning π/2 < θ/2 < 3π/4 (QII).
In QII, cosine is negative, so cos(θ/2) = −√3/3.
Given cosθ = −4/5 (QII), find sin(2θ) and cos(2θ).
Hint: Find sinθ first (positive in QII), then apply the double-angle formulas.
Find the exact value of cos(π/8).
Hint: Use the half-angle formula for cosine with θ = π/4. The angle π/8 is in QI, so the sign is positive.
Simplify: cos²(5x) − sin²(5x)
Hint: Recognize this as the double-angle form of cosine: cos²θ − sin²θ = cos(2θ) with θ = 5x.
Rewrite cos⁴x using power-reduction formulas.
Hint: Write cos⁴x = (cos²x)², apply the power-reduction formula for cos²x, then expand and apply it again.
Given sinθ = −5/13 (QIII), find tan(θ/2).
Hint: Use tan(θ/2) = sinθ/(1 + cosθ). Find cosθ first (negative in QIII).
Interactive Practice — 5 Questions
sin(2θ) equals:
Which form of cos(2θ) is most useful when you know sinθ?
The power-reduction formula for sin²θ is:
If sinθ = 1/2 (QI), then sin(2θ) =
The sign in the half-angle formula sin(θ/2) = ±√((1 − cosθ)/2) is determined by:
Common Mistakes
Writing sin(2θ) = 2sinθ (forgetting the cosθ factor).
sin(2θ) = 2sinθcosθ. Both factors are required.
Using the wrong form of cos(2θ). For example, using cos²θ − sin²θ when only sinθ is known.
Choose the form that matches what you know: use 1 − 2sin²θ when you know sinθ, and 2cos²θ − 1 when you know cosθ.
Determining the ± sign in a half-angle formula based on the quadrant of θ instead of θ/2.
The sign is determined by the quadrant of θ/2. If θ = 270° (QIII), then θ/2 = 135° (QII), where cosine is negative.
Forgetting that the power-reduction formula uses 2θ (double the angle): sin²θ = (1 − cos(2θ))/2.
The argument inside cosine is 2θ, not θ. This is a very common error.
Math Tips
The three forms of cos(2θ) are all equivalent — choose the one that eliminates the unknown. If you know sinθ, use 1 − 2sin²θ. If you know cosθ, use 2cos²θ − 1.
Power-reduction formulas are the key to integrating sin²x and cos²x in calculus. Memorize them now and you will save hours later.
For half-angle problems, always determine the quadrant of θ/2 before choosing the ± sign — not the quadrant of θ.
The identity 2sinθcosθ = sin(2θ) works in reverse too: whenever you see a product of sinθ and cosθ, you can collapse it into a single sine function.
tan(θ/2) = sinθ/(1 + cosθ) = (1 − cosθ)/sinθ — both forms are equivalent. Use whichever avoids a zero denominator.