Deriving the Power-Reduction Identities

Why reduce an even power?

An expression such as \(\sin^4 x\) or \(\cos^6 x\) is a power of a trigonometric function, but it can also be written as a sum of simpler terms involving cosines of multiple angles. The key first step is to rewrite a square using the double-angle identity. Repeating that idea, or expanding with the binomial theorem, reveals the structure for higher even powers.

These identities are useful in trigonometric integration because sums of multiple-angle cosines can be easier to handle than powers. Here the focus is on deriving and understanding the identities themselves, not on carrying out those integrals.

Deriving the half-angle identities

The cosine double-angle identity is \(\cos(2x)=1-2\sin^2 x\). Solving for \(\sin^2 x\) gives the power-reduction identity for sine. The equivalent identity \(\cos(2x)=2\cos^2 x-1\) gives the one for cosine.

These are often called half-angle identities because replacing \(x\) by \(x/2\) expresses the square of a function of the half-angle in terms of the cosine of the full angle. For example, \(\sin^2(x/2)=(1-\cos x)/2\). The same formulas are also called power-reduction identities because they reduce a squared power to a constant and a first power of a cosine.

Fourth powers: applying the identities again

For a fourth power, square the appropriate half-angle identity. Then use the same identity again, this time at angle \(2x\), to rewrite the squared cosine term. The resulting expression contains a constant term and cosines at angles \(2x\) and \(4x\).

Why only multiple-angle cosines appear

The fourth-power examples illustrate the pattern, but a binomial expansion explains it for every nonnegative integer \(n\). Use the complex exponential identities \(e^{ix}=\cos x+i\sin x\) and \(e^{-ix}=\cos x-i\sin x\). Adding them gives \(2\cos x=e^{ix}+e^{-ix}\). Thus \(\cos^{2n}x\) is a binomial power of two exponentials. Terms with opposite exponents combine into cosines, and the exponents are even multiples of \(x\).

Here \(\binom{2n}{k}\) is a binomial coefficient. The terms in either sum have angles \(2x,4x,\ldots,2nx\); the constant term comes from the middle term in the binomial expansion. The sign difference in the sine formula matters: sine's exponential representation contains a subtraction rather than an addition.

Derivation of the general cosine expansion

To see why the theorem holds, expand \(\cos^{2n}x\) using \(2\cos x=e^{ix}+e^{-ix}\). In the expansion, choosing \(k\) factors of \(e^{-ix}\) and \(2n-k\) factors of \(e^{ix}\) produces exponent \(i(2n-2k)x\), with coefficient \(\binom{2n}{k}\).

The sine expansion follows by the same pairing, starting from \(\sin x=(e^{ix}-e^{-ix})/(2i)\). In the \(2n\)-th power, the coefficient of the term with exponent \(i(2n-2k)x\) includes the sign \((-1)^k\), and the common factor is \((-1)^n/2^{2n}\). Pairing the terms indexed by \(k\) and \(2n-k\) gives the cosine coefficient sign \((-1)^{n-k}\), producing the stated sine formula. Thus the finite cosine sum is a consequence of the binomial theorem and the pairing of opposite exponential terms.

Sixth powers from the general pattern

A common sign mistake

The most frequent error is giving the sine-square identity the plus sign used for cosine. For example, the incorrect calculation \(\sin^2 x=(1+\cos(2x))/2\) would give \(\sin^2 0=(1+1)/2=1\), even though \(\sin 0=0\). The correction is to keep track of which double-angle equation is being rearranged:

whereas \(\cos(2x)=2\cos^2 x-1\) gives the plus sign for \(\cos^2 x\). A second useful check is \(x=0\): the sine-square expression must be zero and the cosine-square expression must be one.

What the expansions tell us

Every even power becomes a finite cosine sum with a constant term and even multiples of the original angle. The largest angle is \(2nx\) for a power \(2n\), and the binomial coefficients determine the weights. For sine, alternating signs appear; for cosine, all coefficients in this form are positive.

These are identities, so they hold for every real \(x\), not only for angles where the functions are positive. That matters because even powers are nonnegative, while the cosine terms in their expansions can take either sign. The sum as a whole still equals the original nonnegative power.