10E, 10G - Double-angle identities

10E, 10G - Double-angle identities

Doubling an angle does not simply double its sine or cosine. In 10G, you will set the two angles equal in the supplied addition formulae; in 10E, you will use the Pythagorean identity to rewrite the resulting cosine expression. These connections give several useful forms for exact values and short simplifications.

Deriving double-angle forms

Why is sin⁡60∘\sin60^\circ not twice sin⁡30∘\sin30^\circ? The two values are 32\frac{\sqrt3}{2} and 12\frac12, so simple doubling would incorrectly give 11.

Write the doubled angle as 2θ=θ+θ2\theta=\theta+\theta. The addition formulae then use the same angle twice. This is a double-angle identity: a relationship between a trigonometric ratio at 2θ2\theta and ratios at θ\theta.

Worked example 1: doubling sine and cosine

Question. Use the addition formulae to express sin⁡2x\sin2x and cos⁡2x\cos2x in terms of sin⁡x\sin x and cos⁡x\cos x.

Plan. Replace the two angles in each sum formula by xx, then collect equal products.

sin⁡2x=sin⁡(x+x)write equal angles=sin⁡xcos⁡x+cos⁡xsin⁡xsine addition formula=2sin⁡xcos⁡xcollect equal termscos⁡2x=cos⁡(x+x)write equal angles=cos⁡xcos⁡x−sin⁡xsin⁡xcosine addition formula=cos⁡2x−sin⁡2xwrite each square\begin{aligned} \sin2x&=\sin(x+x) &&\text{write equal angles}\\ &=\sin x\cos x+\cos x\sin x &&\text{sine addition formula}\\ &=2\sin x\cos x &&\text{collect equal terms}\\[4pt] \cos2x&=\cos(x+x) &&\text{write equal angles}\\ &=\cos x\cos x-\sin x\sin x &&\text{cosine addition formula}\\ &=\cos^2x-\sin^2x &&\text{write each square} \end{aligned}

Answer. sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x and cos⁡2x=cos⁡2x−sin⁡2x\cos2x=\cos^2x-\sin^2x, for every real xx.

Check. At x=30∘x=30^\circ, the sine expression is 2×12×32=32=sin⁡60∘2\times\frac12\times\frac{\sqrt3}{2}=\frac{\sqrt3}{2}=\sin60^\circ. The cosine expression is 34−14=12=cos⁡60∘\frac34-\frac14=\frac12=\cos60^\circ.

The equal-angle substitution stays the same when the function changes to tangent. Its quotient introduces restrictions.

Worked example 2: doubling tangent

Question. Use a supplied addition formula to show that tan⁡2x=2tan⁡x1−tan⁡2x\tan2x=\frac{2\tan x}{1-\tan^2x}, stating when this form can be used.

Plan. Set both angles equal to xx in the tangent sum formula.

tan⁡2x=tan⁡(x+x)write the doubled angle=tan⁡x+tan⁡x1−tan⁡xtan⁡xtangent addition formula=2tan⁡x1−tan⁡2xcollect and write the square\begin{aligned} \tan2x&=\tan(x+x) &&\text{write the doubled angle}\\ &=\frac{\tan x+\tan x}{1-\tan x\tan x} &&\text{tangent addition formula}\\ &=\frac{2\tan x}{1-\tan^2x} &&\text{collect and write the square} \end{aligned}

Answer. The form holds for cos⁡x≠0\cos x\ne0 and 1−tan⁡2x≠01-\tan^2x\ne0. These ensure that tan⁡x\tan x and tan⁡2x\tan2x are defined.

Check. If tan⁡x=13\tan x=\frac13, the result is 2/31−1/9=34\frac{2/3}{1-1/9}=\frac34. Independently, a right triangle for an acute such xx gives sin⁡x=110\sin x=\frac1{\sqrt{10}} and cos⁡x=310\cos x=\frac3{\sqrt{10}}. The previous example gives sin⁡2x=610\sin2x=\frac6{10} and cos⁡2x=810\cos2x=\frac8{10}, whose quotient is 34\frac34.

At x=45∘x=45^\circ, the tangent denominator becomes zero and tan⁡90∘\tan90^\circ is undefined. At x=90∘x=90^\circ, tan⁡180∘\tan180^\circ exists, but this particular right-hand expression cannot be used because it contains the undefined tan⁡90∘\tan90^\circ.

Double-angle forms

sin⁡2θ=2sin⁡θcos⁡θ,cos⁡2θ=cos⁡2θ−sin⁡2θ\sin2\theta=2\sin\theta\cos\theta, \qquad \cos2\theta=\cos^2\theta-\sin^2\theta

for every real θ\theta. Also,

tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}

where cos⁡θ≠0\cos\theta\ne0 and 1−tan⁡2θ≠01-\tan^2\theta\ne0.

Derived from the supplied addition formulae; these forms are not printed separately on the formulae sheet.

The input angle can itself be an expression. To expand sin⁡4x\sin4x, double the angle 2x2x.

Equivalent cosine forms

The cosine double-angle identity has two useful alternatives, each involving only one squared ratio.

If you know only sin⁡θ\sin\theta, finding cos⁡θ\cos\theta first may be unnecessary. Since cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta,

cos⁡2θ=(1−sin⁡2θ)−sin⁡2θ=1−2sin⁡2θ.\cos2\theta=(1-\sin^2\theta)-\sin^2\theta=1-2\sin^2\theta.

Alternatively, replace sin⁡2θ\sin^2\theta by 1−cos⁡2θ1-\cos^2\theta:

cos⁡2θ=cos⁡2θ−(1−cos⁡2θ)=2cos⁡2θ−1.\cos2\theta=\cos^2\theta-(1-\cos^2\theta)=2\cos^2\theta-1.

The brackets matter: subtracting 1−cos⁡2θ1-\cos^2\theta adds the final squared term.

Equivalent cosine double-angle forms

For every real θ\theta,

cos⁡2θ=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ.\cos2\theta =\cos^2\theta-\sin^2\theta =2\cos^2\theta-1 =1-2\sin^2\theta.

These are equal expressions; choose one using the ratio you have.

Derived from supplied addition formulae and the Pythagorean identity; not printed separately on the formulae sheet.

No quadrant is needed to calculate cos⁡2θ\cos2\theta from a known sine or cosine alone: these forms use its square. A quadrant would still matter if a calculation required an unknown unsquared ratio.

Worked example 3: a given cosine

Question. Given cos⁡θ=35\cos\theta=\frac35, find cos⁡2θ\cos2\theta exactly.

Plan. Use 2cos⁡2θ−12\cos^2\theta-1, which needs only the given ratio.

cos⁡2θ=2(35)2−1choose the cosine-only form=1825−2525square before multiplying=−725\begin{aligned} \cos2\theta&=2\left(\frac35\right)^2-1 &&\text{choose the cosine-only form}\\ &=\frac{18}{25}-\frac{25}{25} &&\text{square before multiplying}\\ &=-\frac7{25} \end{aligned}

Answer. cos⁡2θ=−725\cos2\theta=-\frac7{25}.

Check. The Pythagorean identity gives sin⁡2θ=1625\sin^2\theta=\frac{16}{25}. The original double-angle form then gives 925−1625=−725\frac9{25}-\frac{16}{25}=-\frac7{25}, agreeing without a choice of sine sign.

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