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Instrument MI-01-449 · Mathematics

Power Reducing Calculator

One doubled angle, two squares removed: enter θ and this sheet returns sin²θ and cos²θ from a single cosine of 2θ, always summing to exactly 1.

Instrument MI-01-449
Sheet 1 OF 1
Rev A
Verified
Type 05 — Trigonometry SER. 2026-01449

sin²(θ)

0.25000000

sin²θ = (1 − cos2θ) ⁄ 2

0.75000000 cos²(θ)
The working Every figure verified twice
  1. sin2 = (1 − cos(2·0.523599)) ⁄ 2 = 0.25000000
  2. cos2 = (1 + cos(2·0.523599)) ⁄ 2 = 0.75000000
Worksheet log
  1. No entries yet — change an input to log a scenario.

How this instrument works

Power-reducing formulas trade a square for a factor of two: sin²θ = (1 − cos2θ) ⁄ 2 and cos²θ = (1 + cos2θ) ⁄ 2 rewrite a squared single-angle trig function as a plain, unsquared cosine evaluated at double the angle. The name describes exactly what happens to the exponent — it drops from 2 down to 1 — while a 2 shows up instead inside the cosine's argument. The angle itself never changes; only the shape of the expression does.

The identity is not a separate law to memorize — it is the double-angle cosine identity read backwards. Start from cos2θ = cos²θ − sin²θ, substitute the Pythagorean identity sin²θ = 1 − cos²θ, and the expression collapses to cos2θ = 2cos²θ − 1, which rearranges directly to cos²θ = (1 + cos2θ) ⁄ 2. Substitute the other way instead, cos²θ = 1 − sin²θ, and that same double-angle identity collapses to cos2θ = 1 − 2sin²θ, giving sin²θ = (1 − cos2θ) ⁄ 2. One relationship, two Pythagorean substitutions, two rearranged formulas.

The formulas earn their keep in calculus, where sin²θ and cos²θ have no elementary antiderivative in their squared form. Swap in (1 − cos2θ) ⁄ 2 or (1 + cos2θ) ⁄ 2 and an integral splits into two ordinary terms, one constant and one plain cosine, both routine to integrate. A related consequence sits inside the formula itself: since cos2θ oscillates evenly between −1 and 1, its average over a full cycle is zero, so sin²θ and cos²θ each average to exactly one half — a fact used directly wherever a squared sinusoid needs averaging over time.

sin2θ=1cos2θ2\sin^2\theta = \frac{1-\cos 2\theta}{2}cos2θ=1+cos2θ2\cos^2\theta = \frac{1+\cos 2\theta}{2}sin2θ+cos2θ=1\sin^2\theta+\cos^2\theta=1
θ — the entered angle · sin²θ, cos²θ — the squared sine and cosine, each rebuilt from a single evaluation of cos 2θ, the cosine of double the angle.
  • Type your angle into the 'Angle, θ' field — the selector beside it defaults to degrees but also reads radians or turns.
  • Internally the sheet forms 2θ and evaluates its cosine once; that single value feeds both output fields below it.
  • 'sin²(θ)' reports (1 − cos 2θ) ⁄ 2, the squared sine rebuilt from that doubled-angle cosine rather than squared directly.
  • 'cos²(θ)' reports (1 + cos 2θ) ⁄ 2 from the same cos 2θ, so the two output fields always sum to exactly 1.
  • Switch the unit selector to radians or turns to see the identical angle in another form without retyping it.

Worked example — θ = 30°, the identity's cleanest check

Set 'Angle, θ' to 30°, which the sheet stores internally as θ = 0.5235987755982988 radians. Doubling it gives 2θ = 60°, where cos 60° is the well-known value 0.5. Plugging that into the formula, sin²(30°) = (1 − 0.5) ⁄ 2 = 0.25, and the field 'sin²(θ)' reports 0.24999999999999994 — matching 0.25 to fifteen decimal places, the tiny residue coming only from binary floating-point representation, never from any error in the identity itself.

The companion field 'cos²(θ)' reports (1 + 0.5) ⁄ 2 = 0.75 from that same cos 60°. Both figures check directly against the plain squares: sin(30°) = 0.5, and 0.5² is 0.25 exactly; cos(30°) = √3 ⁄ 2 ≈ 0.8660254, and squaring that also gives 0.75. Two unrelated routes — squaring the angle's own sine and cosine, or halving one plus or minus the doubled angle's cosine — land on the identical pair of numbers, which is the entire content of the identity laid bare rather than asserted.

Questions

What do the power-reducing formulas actually do?

They rewrite a squared trig function, sin²θ or cos²θ, as a plain unsquared cosine of double the angle, divided by 2 with a shifted constant. sin²θ = (1 − cos2θ) ⁄ 2 and cos²θ = (1 + cos2θ) ⁄ 2 trade an exponent of 2 for a factor of 2 inside the argument — that trade is exactly what 'power reducing' names: the power drops from 2 to 1, and a doubled angle appears in its place.

Where do these formulas come from?

Both fall out of the double-angle cosine identity cos2θ = 1 − 2sin²θ = 2cos²θ − 1, itself built by substituting the Pythagorean identity sin²θ + cos²θ = 1 into cos2θ = cos²θ − sin²θ. Solving the first form for sin²θ gives (1 − cos2θ) ⁄ 2; solving the second for cos²θ gives (1 + cos2θ) ⁄ 2 — one identity, rearranged two ways.

How is this different from the double-angle identities elsewhere on this site?

Double-angle identities start from a single angle and expand outward into 2θ, as in sin(2θ) = 2sinθcosθ, which multiplies two first-power terms together. Power-reducing formulas run the opposite direction: they start from a squared single-angle expression and fold it down into a plain cosine of 2θ. The two families solve for opposite unknowns of the same underlying relationship.

Why do calculus textbooks reach for this identity?

Because sin²θ and cos²θ have no elementary antiderivative in their squared form, but (1 − cos2θ) ⁄ 2 and (1 + cos2θ) ⁄ 2 integrate term by term immediately: ∫sin²θ dθ becomes θ⁄2 − sin(2θ)⁄4 + C, a routine integral rather than a stuck one. The same swap also shows the average value of sin²θ or cos²θ over a full cycle is exactly one half, since cos2θ averages to zero.

Do sin²θ and cos²θ always add up to 1?

Yes, at every angle, since (1 − cos2θ) ⁄ 2 + (1 + cos2θ) ⁄ 2 = 2 ⁄ 2 = 1 regardless of what cos2θ equals — the cos2θ terms cancel outright. That built-in cancellation is a quick self-check: if the two output fields on this sheet ever failed to sum to 1, it would signal a computation error, not a property of the angle entered.

What is the most common mistake with power-reducing formulas?

Forgetting the factor of 2 inside the cosine argument, writing cosθ where cos2θ belongs. At θ = 30°, that slip gives (1 − cos30°) ⁄ 2 ≈ 0.067 instead of the correct sin²(30°) = 0.25 — nowhere close, because cos30° ≈ 0.866 is a very different number from cos60° = 0.5, and the whole formula depends on doubling the angle before taking its cosine.

References