◀ Back to Note

Tony's Intertwingled Memex • Reference Manual

Trigonometric Power-Reduction Formulas

Formulas & Definitions
Core Definition

The trigonometric power-reduction formulas (also known as the half-angle identities for squares) rewrite second-degree trigonometric expressions into first-degree (linear) trigonometric functions of double the angle:

cos⁡2θ=1+cos⁡(2θ)2,sin⁡2θ=1−cos⁡(2θ)2\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}, \qquad \sin^2 \theta = \frac{1 - \cos(2\theta)}{2}
By reducing powers of trigonometric functions, these identities convert quadratic products that lack chain-rule multipliers into directly integrable linear terms.

Concept Explanation

In single-variable and multivariable calculus, integrating even powers of sine and cosine—such as ∫cos⁡2θ dθ\int \cos^2 \theta \, d\theta or ∫sin⁡2θ dθ\int \sin^2 \theta \, d\theta—cannot be solved by basic uu-substitution because there is no isolated sin⁡θ\sin \theta or cos⁡θ\cos \theta multiplying the differential dθd\theta. The power-reduction identities solve this obstacle by eliminating the power in exchange for doubling the frequency of the argument.

1. Algebraic Derivation from the Cosine Double-Angle Formula

Both power-reduction formulas originate directly from the cosine double-angle identity:

cos⁡(2θ)=cos⁡2θ−sin⁡2θ\cos(2\theta) = \cos^2 \theta - \sin^2 \theta

  1. Deriving the Cosine Reduction Formula:
    Replace sin⁡2θ\sin^2 \theta with 1−cos⁡2θ1 - \cos^2 \theta using the Pythagorean identity:

    cos⁡(2θ)=cos⁡2θ−(1−cos⁡2θ)=2cos⁡2θ−1\cos(2\theta) = \cos^2 \theta - (1 - \cos^2 \theta) = 2\cos^2 \theta - 1
    Solve for cos⁡2θ\cos^2 \theta:
    2cos⁡2θ=1+cos⁡(2θ)  ⟹  cos⁡2θ=1+cos⁡(2θ)22\cos^2 \theta = 1 + \cos(2\theta) \implies \cos^2 \theta = \frac{1 + \cos(2\theta)}{2}

  2. Deriving the Sine Reduction Formula:
    Replace cos⁡2θ\cos^2 \theta with 1−sin⁡2θ1 - \sin^2 \theta:

    cos⁡(2θ)=(1−sin⁡2θ)−sin⁡2θ=1−2sin⁡2θ\cos(2\theta) = (1 - \sin^2 \theta) - \sin^2 \theta = 1 - 2\sin^2 \theta
    Solve for sin⁡2θ\sin^2 \theta:
    2sin⁡2θ=1−cos⁡(2θ)  ⟹  sin⁡2θ=1−cos⁡(2θ)22\sin^2 \theta = 1 - \cos(2\theta) \implies \sin^2 \theta = \frac{1 - \cos(2\theta)}{2}

  3. Tangent Power-Reduction Formula:
    Taking the ratio of the sine and cosine formulas yields:

    tan⁡2θ=sin⁡2θcos⁡2θ=1−cos⁡(2θ)1+cos⁡(2θ)\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{1 - \cos(2\theta)}{1 + \cos(2\theta)}

When to Use It / Method Comparison

Integrand Form Algebraic Clue Recommended Technique Why?
sin⁡mθcos⁡nθ\sin^m \theta \cos^n \theta with at least one odd power e.g., ∫sin⁡3θcos⁡2θ dθ\int \sin^3 \theta \cos^2 \theta \, d\theta uu-Substitution Peel off one odd factor as dudu and convert the remaining even power using sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.
sin⁡mθcos⁡nθ\sin^m \theta \cos^n \theta with both powers even e.g., ∫cos⁡2θ dθ\int \cos^2 \theta \, d\theta, ∫sin⁡2θcos⁡2θ dθ\int \sin^2 \theta \cos^2 \theta \, d\theta Power-Reduction Formulas No standalone odd factor exists to act as dudu. Powers must be systematically reduced to linear cosine terms.
Mixed fraction containing (1±cos⁡θ)(1 \pm \cos\theta) e.g., 11+cos⁡θ\frac{1}{1 + \cos \theta} Conjugate Multiplication or Half-Angle Multiply by conjugate to create single-term denominators via sin⁡2θ\sin^2\theta.

Why We Use It in Calculus (Calculus III Context)

In multivariable calculus (MATH-283), power-reduction formulas are indispensable when evaluating Double Integrals in Polar Coordinates (Session 15) and Triple Integrals in Cylindrical and Spherical Coordinates (Sessions 16–18):

  1. Angular Integrals over Symmetric Domains:
    When calculating areas, mass, or moments of inertia over circular disks or sectors, converting Cartesian coordinates (x=rcos⁡θ,y=rsin⁡θx = r\cos\theta, y = r\sin\theta) generates terms like:
    ∬Dx2 dA=∫02π∫0R(r2cos⁡2θ) r dr dθ=(∫0Rr3 dr)(∫02πcos⁡2θ dθ)\iint_D x^2 \, dA = \int_0^{2\pi} \int_0^R (r^2 \cos^2 \theta) \, r \, dr \, d\theta = \left(\int_0^R r^3 \, dr\right) \left(\int_0^{2\pi} \cos^2 \theta \, d\theta\right)
    Evaluating the angular factor requires power reduction:
    ∫02πcos⁡2θ dθ=∫02π1+cos⁡(2θ)2 dθ=[θ2+sin⁡(2θ)4]02π=2π2+0=π\int_0^{2\pi} \cos^2 \theta \, d\theta = \int_0^{2\pi} \frac{1 + \cos(2\theta)}{2} \, d\theta = \left[ \frac{\theta}{2} + \frac{\sin(2\theta)}{4} \right]_0^{2\pi} = \frac{2\pi}{2} + 0 = \pi
  2. Periodic Cancellation over Complete Rotations:
    Notice the remarkable property: because sin⁡(2θ)\sin(2\theta) is 2π2\pi-periodic, the oscillating term ∫02πcos⁡(2θ) dθ=0\int_0^{2\pi} \cos(2\theta) \, d\theta = 0 collapses to zero over any full rotation. Thus:
    ∫02πcos⁡2θ dθ=πand∫02πsin⁡2θ dθ=π\int_0^{2\pi} \cos^2 \theta \, d\theta = \pi \qquad \text{and} \qquad \int_0^{2\pi} \sin^2 \theta \, d\theta = \pi
  3. Calculating Centroids and Moments of Inertia:
    Computing polar moments of inertia Ix=∬y2 dmI_x = \iint y^2 \, dm and Iy=∬x2 dmI_y = \iint x^2 \, dm in rotating machinery (such as turbine disks and flywheels) requires exact evaluation of sin⁡2θ\sin^2\theta and cos⁡2θ\cos^2\theta integral factors.

Example Problems

Example Problem 1: Definite Integral of an Even Power

Evaluate the definite integral:

∫0πsin⁡2θ dθ\int_{0}^{\pi} \sin^2 \theta \, d\theta

Solution

Step 1: Apply the power-reduction identity
Substitute sin⁡2θ=1−cos⁡(2θ)2\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}:

∫0πsin⁡2θ dθ=∫0π1−cos⁡(2θ)2 dθ=12∫0π(1−cos⁡(2θ)) dθ\int_{0}^{\pi} \sin^2 \theta \, d\theta = \int_{0}^{\pi} \frac{1 - \cos(2\theta)}{2} \, d\theta = \frac{1}{2} \int_{0}^{\pi} \Big(1 - \cos(2\theta)\Big) \, d\theta

Step 2: Find the antiderivative
Integrate term-by-term:

12[θ−12sin⁡(2θ)]0π\frac{1}{2} \left[ \theta - \frac{1}{2}\sin(2\theta) \right]_{0}^{\pi}

Step 3: Evaluate at the limits

=12[(π−12sin⁡(2π))−(0−12sin⁡(0))]= \frac{1}{2} \left[ \left(\pi - \frac{1}{2}\sin(2\pi)\right) - \left(0 - \frac{1}{2}\sin(0)\right) \right]
Since sin⁡(2π)=0\sin(2\pi) = 0 and sin⁡(0)=0\sin(0) = 0:
=12[π−0]=π2= \frac{1}{2} [\pi - 0] = \frac{\pi}{2}

Answer:
π2\frac{\pi}{2}

Example Problem 2: Polar Double Integral Involving Trigonometric Powers

Evaluate the double integral over the disk D={(x,y)∈R2∣x2+y2≤4}D = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 \le 4\}:

∬Dx2 dA\iint_D x^2 \, dA

Solution

Step 1: Convert to polar coordinates
In polar coordinates, the disk DD is bounded by 0≤r≤20 \le r \le 2 and 0≤θ≤2π0 \le \theta \le 2\pi.
Substitute x=rcos⁡θx = r \cos \theta and dA=r dr dθdA = r \, dr \, d\theta:

∬Dx2 dA=∫02π∫02(rcos⁡θ)2 r dr dθ=∫02π∫02r3cos⁡2θ dr dθ\iint_D x^2 \, dA = \int_{0}^{2\pi} \int_{0}^{2} (r \cos \theta)^2 \, r \, dr \, d\theta = \int_{0}^{2\pi} \int_{0}^{2} r^3 \cos^2 \theta \, dr \, d\theta

Step 2: Exploit product separability
Because the integration limits are constants and the integrand factors into independent functions of rr and θ\theta:

=(∫02r3 dr)(∫02πcos⁡2θ dθ)= \left( \int_{0}^{2} r^3 \, dr \right) \left( \int_{0}^{2\pi} \cos^2 \theta \, d\theta \right)

Step 3: Evaluate the radial integral

∫02r3 dr=[r44]02=164−0=4\int_{0}^{2} r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^2 = \frac{16}{4} - 0 = 4

Step 4: Evaluate the angular integral using power reduction
Apply cos⁡2θ=1+cos⁡(2θ)2\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}:

∫02πcos⁡2θ dθ=12[θ+sin⁡(2θ)2]02π=12[(2π+0)−(0+0)]=π\int_{0}^{2\pi} \cos^2 \theta \, d\theta = \frac{1}{2} \left[ \theta + \frac{\sin(2\theta)}{2} \right]_0^{2\pi} = \frac{1}{2} \Big[(2\pi + 0) - (0 + 0)\Big] = \pi

Step 5: Multiply the results

∬Dx2 dA=(4)(π)=4π\iint_D x^2 \, dA = (4)(\pi) = 4\pi

Answer:
4π4\pi

Connections & References