CP 2.8 - Powers of sine and cosine | Edexcel A-Level Further Mathematics

Turn powers of sine and cosine into sums of multiple-angle terms by expanding reciprocal pairs of complex numbers. You will also express powers of tangent using multiple angles and use the expansions to integrate trigonometric powers.

A unit complex number and its reciprocal

What is the exact area under $y = \cos^4 x$ from $x = 0$ to $x = \frac{\pi}{2}$ ? Directly integrating a fourth power is awkward. We will replace it by a constant and two ordinary cosine terms; then the area takes just a few lines.

You already know

De Moivre connects $z^m$ with a multiple angle when $z = \cos\theta + \mathrm{i}\sin\theta$ . The preceding lesson expanded $z^m$ to obtain multiple-angle formulae. Here we reverse the direction: a power of $\cos\theta$ or $\sin\theta$ will become a sum of multiple-angle terms.

For $z = \cos\theta + \mathrm{i}\sin\theta$ , the modulus is $1$ . Its reciprocal is its conjugate because $zz^* = 1$ . For example, at $\theta = \frac{\pi}{3}$ ,

\begin{aligned} z &= \frac12 + \frac{\mathrm{i}\sqrt{3}}2 \\ z^{-1} &= \frac12 - \frac{\mathrm{i}\sqrt{3}}2 \\ z+z^{-1} &= 1 = 2\cos\frac{\pi}{3} \\ z-z^{-1} &= \mathrm{i}\sqrt{3} = 2\mathrm{i}\sin\frac{\pi}{3} \end{aligned}

Adding cancels the imaginary components; subtracting cancels the real ones. The same cancellation works with every integer power.

Reciprocal pairs

For real $\theta$ , $z = \cos\theta + \mathrm{i}\sin\theta$ and any integer $m$ ,

z^m + z^{-m} = 2\cos m\theta \qquad z^m - z^{-m} = 2\mathrm{i}\sin m\theta

These follow from de Moivre and the evenness of cosine and oddness of sine. They require the chosen $z$ to have modulus $1$ .

Not in the formula booklet: learn it.

Indeed, $z^{-m}=\cos m\theta-\mathrm{i}\sin m\theta$ , so the cancellation is visible before any expansion. The restriction matters: for $z=2$ , $z+z^{-1}=\frac52$ , whereas twice the cosine of its angle is only $2$ .

Key idea

Pair opposite powers: a sum gives cosine and a difference gives sine. Check that the complex number has modulus $1$ before using these identities.

Choose the expression that preserves both the factor $2$ and the imaginary unit.

Expanding powers of cosine

Start from $z+z^{-1}=2\cos\theta$ , raise both sides to the required power and expand. Then group opposite powers before replacing them with cosines.

If the exponent is odd, the expansion contains the pairs $z^m+z^{-m}$ with odd $m$ . If the exponent is even, the pairs have even $m$ , and there is a constant term. This gives a quick check on the angles you should see at the end.

A cosine power to expand

For $z = \cos\theta+\mathrm{i}\sin\theta$ and a positive integer $n$ ,

(z+z^{-1})^n=2^n\cos^n\theta

Expand the left side and use $z^m+z^{-m}=2\cos m\theta$ . Power-to-multiple-angle identities derived this way are not separately printed.

Not in the formula booklet: learn it.

The cubic case makes the grouping particularly clear.

Worked example 1: a cubed cosine

Question. Express $\cos^3\theta$ in terms of $\cos 3\theta$ and $\cos\theta$ , using de Moivre's theorem.

Plan. Expand $(z+z^{-1})^3$ , pair opposite powers and divide by $8$ .

\begin{aligned} 8\cos^3\theta &= (z+z^{-1})^3 && \text{cube }2\cos\theta \\ &= z^3+3z+3z^{-1}+z^{-3} && \text{binomial expansion} \\ &= (z^3+z^{-3})+3(z+z^{-1}) && \text{pair opposite powers} \\ &= 2\cos 3\theta+6\cos\theta && \text{reciprocal-pair identities} \end{aligned}

Why this step? Both terms in each pair have the same coefficient. Grouping them exposes exactly the expressions de Moivre converts to cosine.

Answer. $\displaystyle \cos^3\theta=\frac{\cos 3\theta+3\cos\theta}{4}$ .

Check. At $\theta=\frac{\pi}{3}$ , the expression gives $\frac{-1+3/2}{4}=\frac18=(\frac12)^3$ .

Change only the exponent from $3$ to $5$ . The outer powers differ by two because each factor of $z^{-1}$ replaces a factor of $z$ .

Worked example 2: a fifth power in exam style

Question. Use de Moivre's theorem to show that

\cos^5\theta=\frac{\cos 5\theta+5\cos 3\theta+10\cos\theta}{16}

Plan. Expand the fifth power of $z+z^{-1}$ and keep all three reciprocal pairs visible.

\begin{aligned} 32\cos^5\theta &= (z+z^{-1})^5 && \text{fifth power} \\ &= z^5+5z^3+10z+10z^{-1}+5z^{-3}+z^{-5} && \text{expand and simplify indices} \\ &= (z^5+z^{-5})+5(z^3+z^{-3})+10(z+z^{-1}) && \text{group pairs} \\ &= 2\cos 5\theta+10\cos 3\theta+20\cos\theta && \text{de Moivre pairs} \end{aligned}

Why this step? The binomial middle terms are $10z^3(z^{-1})^2=10z$ and $10z^2(z^{-1})^3=10z^{-1}$ . Simplifying those indices before grouping avoids a common wrong angle.

Answer. Dividing by $32$ gives the printed result, with denominator $16$ .

Check. At $\theta=0$ , the right side is $\frac{1+5+10}{16}=1=\cos^5 0$ .

The final conversion has another factor of $2$ in every reciprocal pair, so the denominator is $16$ , not $32$ .

Watch out

Wrong: $(z+z^{-1})^5=2\cos^5\theta$ . Right: $(2\cos\theta)^5=32\cos^5\theta$ . The power applies to the $2$ as well as to the cosine.

For an even power, the middle binomial term contributes a constant. Show it rather than trying to fit it into a pair.

Expanding powers of sine

For sine, use $z-z^{-1}=2\mathrm{i}\sin\theta$ . Two signs now need care: the alternating binomial signs and the power of $\mathrm{i}$ on the other side.

For example, $(2\mathrm{i})^3=-8\mathrm{i}$ , whereas $(2\mathrm{i})^4=16$ . Work out this factor before simplifying the trigonometric expression; it stops a sign error from spreading through the proof.

A sine power to expand

For $z=\cos\theta+\mathrm{i}\sin\theta$ and a positive integer $n$ ,

(z-z^{-1})^n=(2\mathrm{i})^n\sin^n\theta

After expansion, odd powers group into differences and give sines; even powers group into sums and give cosines. The derived sine-power identities are not printed separately.

Not in the formula booklet: learn it.

The cubic example shows why an overall negative sign appears before it disappears into the rearrangement.

Worked example 3: a cubed sine

Question. Express $\sin^3\theta$ in terms of multiple-angle sines.

Plan. Cube $z-z^{-1}$ , group differences and divide by $-8\mathrm{i}$ .

\begin{aligned} -8\mathrm{i}\sin^3\theta &=(z-z^{-1})^3 && (2\mathrm{i})^3=-8\mathrm{i} \\ &=z^3-3z+3z^{-1}-z^{-3} && \text{alternating signs} \\ &=(z^3-z^{-3})-3(z-z^{-1}) && \text{group differences} \\ &=2\mathrm{i}\sin 3\theta-6\mathrm{i}\sin\theta && \text{de Moivre pairs} \end{aligned}

Why this step? The pair $-3z+3z^{-1}$ is $-3(z-z^{-1})$ . Factoring out the negative coefficient preserves both signs.

Answer. $\displaystyle \sin^3\theta=\frac{3\sin\theta-\sin 3\theta}{4}$ .

Check. At $\theta=\frac{\pi}{6}$ , the expression is $\frac{3/2-1}{4}=\frac18=\sin^3\frac{\pi}{6}$ .

Change only the exponent from $3$ to $4$ . The even power makes the final expression real before division, so it uses cosine pairs.

Worked example 4: a fourth power of sine

Question. Use de Moivre's theorem to show that $8\sin^4\theta=3-4\cos 2\theta+\cos 4\theta$ .

Plan. Expand the fourth power, pair opposite powers and use $(2\mathrm{i})^4=16$ .

\begin{aligned} 16\sin^4\theta &=(z-z^{-1})^4 && \mathrm{i}^4=1 \\ &=z^4-4z^2+6-4z^{-2}+z^{-4} && \text{binomial expansion} \\ &=(z^4+z^{-4})-4(z^2+z^{-2})+6 && \text{group sums} \\ &=2\cos 4\theta-8\cos 2\theta+6 && \text{convert pairs} \end{aligned}

Why this step? Both terms in the second pair have coefficient $-4$ . Writing the bracket explicitly is what justifies the $-8\cos 2\theta$ term.

Answer. Dividing by $2$ gives $8\sin^4\theta=3-4\cos 2\theta+\cos 4\theta$ .

Check. At $\theta=\frac{\pi}{6}$ , the right side is $3-2-\frac12=\frac12=8(\frac12)^4$ .

The grouping is part of the argument. In a show-that question, a jump from an ungrouped expansion to the printed answer can hide the step you need to demonstrate.

Watch out

Wrong: $(2\mathrm{i}\sin\theta)^4=-16\sin^4\theta$ . Right: $\mathrm{i}^4=1$ , so the result is $+16\sin^4\theta$ . For odd powers, retain the factor of $\mathrm{i}$ until both sides can be divided by it.

The next odd exponent gives three difference pairs rather than two.

Powers of tangent as multiple-angle ratios

Divide a sine-power identity by the corresponding cosine-power identity to obtain a power of tangent. You normally get a ratio of multiple-angle expressions, rather than a finite sum.

The original restriction is $\cos\theta \ne 0$ . Preserve it even if later algebra cancels a factor. The square and fourth-power cases illustrate the method without introducing any new complex-number rule.

Forming tangent powers

For a positive integer $n$ and $\cos\theta \ne 0$ ,

\tan^n\theta=\frac{\sin^n\theta}{\cos^n\theta}

Substitute the two multiple-angle expansions. Their common numerical factor cancels.

Not in the formula booklet: learn it.

For the square, the expansions of $(z+z^{-1})^2$ and $(z-z^{-1})^2$ give $2\cos^2\theta=1+\cos 2\theta$ and $2\sin^2\theta=1-\cos 2\theta$ . These double-angle rearrangements are not separately printed in the formula booklet.

Worked example 5: a squared tangent

Question. Express $\tan^2\theta$ in terms of $\cos 2\theta$ .

Plan. Divide the sine-square expansion by the cosine-square expansion.

\begin{aligned} \tan^2\theta &=\frac{\sin^2\theta}{\cos^2\theta} && \cos\theta\ne0 \\ &=\frac{(1-\cos 2\theta)/2}{(1+\cos 2\theta)/2} && \text{square expansions} \\ &=\frac{1-\cos 2\theta}{1+\cos 2\theta} && \text{cancel the common factor} \end{aligned}

Why this step? The numerator and denominator must come from the same power. Dividing a sine-square identity by a cosine fourth-power identity would not produce $\tan^2\theta$ .

Answer. $\displaystyle \tan^2\theta=\frac{1-\cos 2\theta}{1+\cos 2\theta}$ , for $\cos\theta\ne0$ .

Check. At $\theta=\frac{\pi}{4}$ , the fraction is $1$ , agreeing with $\tan^2\frac{\pi}{4}=1$ .

Change only the exponent from $2$ to $4$ . The final expression now needs the angles $2\theta$ and $4\theta$ .

Worked example 6: a fourth power of tangent

Question. Express $\tan^4\theta$ in terms of $\cos 2\theta$ and $\cos 4\theta$ .

Plan. Divide the two fourth-power identities, cancelling their common denominator $8$ .

\begin{aligned} \tan^4\theta &=\frac{\sin^4\theta}{\cos^4\theta} && \text{same power} \\ &=\frac{(3-4\cos 2\theta+\cos 4\theta)/8}{(3+4\cos 2\theta+\cos 4\theta)/8} && \text{fourth-power expansions} \\ &=\frac{3-4\cos 2\theta+\cos 4\theta}{3+4\cos 2\theta+\cos 4\theta} && \text{cancel }1/8 \end{aligned}

Why this step? The denominator equals $8\cos^4\theta$ , so it vanishes exactly where tangent itself is undefined. The restriction is retained automatically.

Answer. The displayed ratio equals $\tan^4\theta$ for $\cos\theta\ne0$ .

Check. At $\theta=\frac{\pi}{4}$ , numerator and denominator are both $2$ , so the ratio is $1$ .

These two derived tangent-power identities are not printed in the formula booklet; you can reconstruct them from the paired expansions.

Watch out

Wrong: a formula for $\tan^4\theta$ is a formula for $\tan 4\theta$ . The first raises a value to a power; the second changes the angle. Check the position of the $4$ before choosing a method.

For an odd power, divide the odd-power sine and cosine expansions instead.

Integrating after the expansion

The point of lowering a power is often to make an integral routine. Expand first, divide by the identity's numerical factor, then integrate each constant or multiple-angle term.

You already know

From A-level Mathematics, $\int\cos(kx)\,\mathrm{d}x=\frac1k\sin(kx)+C$ for non-zero constant $k$ . A definite integral is the antiderivative at the upper limit minus its value at the lower limit. These basic integration rules are not printed in the formula booklet; they are assumed knowledge here.

The denominator from the power expansion remains attached to every term. Keep it outside the integral until you have integrated the bracket.

Worked example 7: integrating a fourth power of cosine

Question. Find $\int\cos^4 x\,\mathrm{d}x$ .

Plan. Use $8\cos^4 x=3+4\cos 2x+\cos 4x$ , then integrate term by term.

\begin{aligned} \int\cos^4 x\,\mathrm{d}x &=\frac18\int(3+4\cos 2x+\cos 4x)\,\mathrm{d}x && \text{power expansion} \\ &=\frac18\left(3x+2\sin 2x+\frac14\sin 4x\right)+C && \text{divide by each frequency} \\ &=\frac{3x}{8}+\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+C && \text{simplify coefficients} \end{aligned}

Why this step? Integrating $\cos 4x$ introduces a factor $\frac14$ , in addition to the outer $\frac18$ ; together they give $\frac1{32}$ .

Answer. $\displaystyle \frac{3x}{8}+\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+C$ .

Check. Differentiating gives $\frac38+\frac12\cos 2x+\frac18\cos 4x=\cos^4 x$ .

Change only cosine to sine, keeping the fourth power. The sign of the double-angle term changes; its integration factor does not.

Worked example 8: integrating a fourth power of sine

Question. Find $\int\sin^4 x\,\mathrm{d}x$ .

Plan. Use the sine fourth-power expansion and integrate the same three types of term.

\begin{aligned} \int\sin^4 x\,\mathrm{d}x &=\frac18\int(3-4\cos 2x+\cos 4x)\,\mathrm{d}x && \text{sine expansion} \\ &=\frac18\left(3x-2\sin 2x+\frac14\sin 4x\right)+C && \text{integrate each term} \\ &=\frac{3x}{8}-\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+C && \text{simplify} \end{aligned}

Why this step? Using the cosine expansion by mistake would preserve all the integration working but give the wrong sign in the answer. Identify the original function before substituting.

Answer. $\displaystyle \frac{3x}{8}-\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+C$ .

Check. Differentiating gives $\frac38-\frac12\cos 2x+\frac18\cos 4x=\sin^4 x$ .

These antiderivatives are consequences of the power identities and are not printed in the formula booklet. Now the opening area follows from Worked example 7:

\int_0^{\pi/2}\cos^4 x\,\mathrm{d}x =\left[\frac{3x}{8}+\frac{\sin 2x}{4}+\frac{\sin 4x}{32}\right]_0^{\pi/2} =\frac{3\pi}{16}

The oscillating terms also integrate to zero over a complete turn. Consequently, the exact mean value of $\cos^4 x$ over $0\le x\le2\pi$ is $\frac38$ : it is the constant term in the expansion. A complex-number calculation has revealed an average height on a real graph.

Watch out

Wrong: $\int\cos 4x\,\mathrm{d}x=\sin 4x$ . Right: $\frac14\sin 4x$ , since differentiating the sine produces another factor of $4$ . Keep both the frequency factor and the expansion's outer factor.

Changing the angle inside the original fourth power changes both frequencies in the expansion.

Quick Recap

For a unit-modulus $z$ , opposite powers add to give $2\cos m\theta$ and subtract to give $2\mathrm{i}\sin m\theta$ .
Expand $(z+z^{-1})^n$ for cosine powers and $(z-z^{-1})^n$ for sine powers; simplify the powers and group them explicitly.
Keep $2^n$ , $\mathrm{i}^n$ and the binomial signs separate until they are settled.
Divide matching sine- and cosine-power identities for tangent powers, retaining $\cos\theta\ne0$ .
In integration, keep the expansion factor and divide by each new frequency.

The opening area is $\frac{3\pi}{16}$ ; the fourth power becomes a constant plus two cosines whose endpoint sine terms vanish.

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CP 2.8 - Powers of sine and cosine

A unit complex number and its reciprocal

Expanding powers of cosine

Expanding powers of sine

Powers of tangent as multiple-angle ratios

Integrating after the expansion

Quick Recap

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