2.5 - Trigonometric powers and sums | Cambridge A-Level Further Mathematics

Complex numbers can combine a sine calculation and a cosine calculation into one expression. You will express powers of sine and cosine as multiple angles, sum trigonometric series by the C plus i S method, and recognise binomial coefficients as the terms of a complex binomial expansion.

Powers into multiple angles

Can $\sin^6\theta$ be rewritten using only cosines of $2\theta$ , $4\theta$ and $6\theta$ ? A sixth power looks large, but a complex number and its reciprocal make the terms pair up neatly.

Let $z=\cos\theta+i\sin\theta$ . As its modulus is $1$ ,

z^{-1}=\cos\theta-i\sin\theta.

Adding and subtracting gives $z+z^{-1}=2\cos\theta$ and $z-z^{-1}=2i\sin\theta$ . De Moivre also gives $z^m+z^{-m}=2\cos m\theta$ , for every integer $m$ .

Conjugate expressions for sine and cosine

With $z=\cos\theta+i\sin\theta$ ,

\cos\theta=\frac{z+z^{-1}}2,\qquad \sin\theta=\frac{z-z^{-1}}{2i},\qquad z^m+z^{-m}=2\cos m\theta.

These formulae are not in the formula booklet: learn how to obtain them. The binomial coefficient formula is supplied.

For example, squaring the first expression gives $\cos^2\theta=(z^2+2+z^{-2})/4=(1+\cos2\theta)/2$ . Higher even powers use the same pairing. The exponent on $\cos^6\theta$ means a sixth power; $\cos6\theta$ means a multiplied angle.

Worked example 1: reducing a sixth power of cosine

Question. Express $\cos^6\theta$ in terms of cosines of multiple angles.

Plan. Expand $(z+z^{-1})^6$ and pair opposite powers of $z$ .

\begin{aligned} 64\cos^6\theta &=z^6+6z^4+15z^2+20+15z^{-2}+6z^{-4}+z^{-6}\\ &=(z^6+z^{-6})+6(z^4+z^{-4})\\ &\quad+15(z^2+z^{-2})+20\\ &=2\cos6\theta+12\cos4\theta+30\cos2\theta+20. \end{aligned}

Pairing creates real cosine terms and removes the imaginary parts.

Answer.

\cos^6\theta=\frac{10+15\cos2\theta+6\cos4\theta+\cos6\theta}{32}.

Check. At $\theta=\pi/6$ , the original power is $(\sqrt3/2)^6=27/64$ . The new expression is $(10+15/2-3-1)/32=27/64$ .

Change only cosine to sine. The alternating signs in the binomial expansion matter, as does $i^6=-1$ in the denominator.

Worked example 2: reducing a sixth power of sine

Question. Express $\sin^6\theta$ using $\cos2\theta$ , $\cos4\theta$ and $\cos6\theta$ .

Plan. Expand $(z-z^{-1})^6$ , account for $(2i)^6=-64$ , and pair the powers.

\begin{aligned} -64\sin^6\theta &=z^6-6z^4+15z^2-20+15z^{-2}-6z^{-4}+z^{-6}\\ &=2\cos6\theta-12\cos4\theta+30\cos2\theta-20. \end{aligned}

Divide by $-64$ ; every sign changes.

Answer.

\sin^6\theta=\frac{10-15\cos2\theta+6\cos4\theta-\cos6\theta}{32}.

Check. At $\theta=\pi/6$ , direct evaluation gives $1/64$ . The formula gives $(10-15/2-3+1)/32=1/64$ . At $\theta=0$ , its numerator is $10-15+6-1=0$ , as required.

The two results answer the opening question. Their constant terms agree, while the $2\theta$ and $6\theta$ coefficients change sign. Such a form is also convenient when integrating a trigonometric power, because each cosine term can be integrated separately.

Watch out

$(2i)^6$ is $-64$ , not $64$ . Losing this sign would make the sine power negative at some angles, contradicting $\sin^6\theta\ge0$ .

Use the same method with a lower even power.

The C plus i S method

If a cosine series and a sine series have the same coefficients and angles, add them as real and imaginary parts of one complex series.

For instance, define

C=\sum_{r=0}^{n-1}\rho^r\cos r\theta,\qquad S=\sum_{r=0}^{n-1}\rho^r\sin r\theta.

Using de Moivre,

C+iS=\sum_{r=0}^{n-1}\bigl(\rho(\cos\theta+i\sin\theta)\bigr)^r.

This is a geometric progression with common ratio $q=\rho(\cos\theta+i\sin\theta)$ . Sum that progression first; take the real and imaginary parts only after simplifying the complex fraction.

The C plus i S method

For the finite sums above,

C+iS=\frac{1-q^n}{1-q}\quad(q\ne1).

If $q=1$ , the sum is $n$ . For an infinite series with $|q|<1$ ,

C+iS=\frac1{1-q}.

The geometric-series formulae are in the formula booklet. Combining the two trigonometric sums into one complex series is the method to learn.

There is also a useful exact form for an unweighted finite sum. Take $\rho=1$ and write $q=e^{i\theta}$ , using $e^{i\theta}=\cos\theta+i\sin\theta$ . Factoring each difference into a half-angle form gives

\begin{aligned} \sum_{r=0}^{n-1}e^{ir\theta} &=\frac{1-e^{in\theta}}{1-e^{i\theta}}\\ &=e^{i(n-1)\theta/2}\frac{\sin(n\theta/2)}{\sin(\theta/2)}. \end{aligned}

Indeed, $1-e^{ix}=-2i e^{ix/2}\sin(x/2)$ ; this explains both the phase and the real multiplier. The expression requires $\theta\ne2k\pi$ , for integers $k$ . At $\theta=2k\pi$ , directly $C=n$ and $S=0$ . The signs of the sine factors must be retained; they are not automatically positive moduli.

Worked example 3: a finite weighted sum

Question. Find the exact values of

C=\sum_{r=0}^{3}\left(\frac1{\sqrt2}\right)^r\cos\frac{r\pi}{4},\qquad S=\sum_{r=0}^{3}\left(\frac1{\sqrt2}\right)^r\sin\frac{r\pi}{4}.

Plan. Combine the sums into a four-term geometric progression with ratio $(1+i)/2$ .

Here $q=(1+i)/2$ , so $q^2=i/2$ and $q^4=-1/4$ . Therefore

\begin{aligned} C+iS&=\frac{1-q^4}{1-q}\\ &=\frac{5/4}{(1-i)/2}\\ &=\frac54(1+i) &&\text{multiply by the conjugate}. \end{aligned}

Answer. $C=5/4$ and $S=5/4$ .

Check. Directly adding the four complex terms gives

1+\frac{1+i}{2}+\frac i2+\frac{-1+i}{4}=\frac54+\frac54i.

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2.5 - Trigonometric powers and sums

Powers into multiple angles

The C plus i S method

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