2.3, 3.3 - Double-angle identities

2.3, 3.3 - Double-angle identities

A double-angle identity connects a function of twice an angle to functions of the original angle. In this shared lesson for 2.3 and 3.3, you will derive the identities, choose the form that suits the information given, and solve equations without losing roots or accepting undefined expressions.

Three forms of cosine

If sin⁡A=3/5\sin A=3/5, can you find cos⁡2A\cos2A without first finding AA? Yes: a double-angle identity uses the supplied ratio directly and keeps the answer exact.

Set B=AB=A in cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B)=\cos A\cos B-\sin A\sin B. This gives cos⁡2A=cos⁡2A−sin⁡2A\cos2A=\cos^2A-\sin^2A. Replacing one squared function using sin⁡2A+cos⁡2A=1\sin^2A+\cos^2A=1 produces two other forms.

Three cosine double-angle forms

For every real angle AA,

cos⁡2A=cos⁡2A−sin⁡2A=2cos⁡2A−1=1−2sin⁡2A.\cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A.

In the formula booklet.

Choose a form that uses what you already know. Given sine, use 1−2sin⁡2A1-2\sin^2A; given cosine, use 2cos⁡2A−12\cos^2A-1. In an equation involving sin⁡x\sin x, the sine form avoids introducing cosine as a second unknown.

Worked example 1: given sine

Question. Given sin⁡A=3/5\sin A=3/5, find the exact value of cos⁡2A\cos2A.

Plan. Choose the sine-only form so no angle or square root is needed.

cos⁡2A=1−2sin⁡2A=1−2(35)2=1−1825=725.\begin{aligned} \cos2A&=1-2\sin^2A\\ &=1-2\left(\frac35\right)^2\\ &=1-\frac{18}{25}=\frac7{25}. \end{aligned}

Answer. cos⁡2A=7/25\cos2A=7/25.

Check. The basic identity gives cos⁡2A=16/25\cos^2A=16/25, so cos⁡2A−sin⁡2A=16/25−9/25=7/25\cos^2A-\sin^2A=16/25-9/25=7/25. Either possible sign of cos⁡A\cos A gives this same answer.

Change only the supplied function from sine to cosine, keeping the ratio 3/53/5. This changes the useful form and reverses the result's sign.

Worked example 2: given cosine

Question. Given cos⁡A=3/5\cos A=3/5, find the exact value of cos⁡2A\cos2A.

Plan. Choose 2cos⁡2A−12\cos^2A-1.

cos⁡2A=2cos⁡2A−1=2(35)2−1=1825−1=−725.\begin{aligned} \cos2A&=2\cos^2A-1\\ &=2\left(\frac35\right)^2-1\\ &=\frac{18}{25}-1=-\frac7{25}. \end{aligned}

Answer. cos⁡2A=−7/25\cos2A=-7/25.

Check. Now sin⁡2A=16/25\sin^2A=16/25, so the difference of squares is 9/25−16/25=−7/259/25-16/25=-7/25. Both answers lie in [−1,1][-1,1], as cosine must.

These calculations do not require a quadrant for AA, because only squares of its ratios are used. A quadrant will matter when an unsquared sine or cosine is needed.

Watch out

cos⁡2A\cos2A means cos⁡(2A)\cos(2A); cos⁡2A\cos^2A means (cos⁡A)2(\cos A)^2. They are different: at A=60∘A=60^\circ, the first is −1/2-1/2, while the second is 1/41/4.

Use the cosine-only form for a fresh ratio.

Sine and tangent

Setting the two angles equal in the sine compound-angle formula gives sin⁡2A=2sin⁡Acos⁡A\sin2A=2\sin A\cos A. Unlike the cosine forms, it uses the signs of both ratios. Dividing this identity by the cosine double-angle identity then gives a tangent form.

Sine and tangent double-angle formulae

sin⁡2A=2sin⁡Acos⁡A\sin2A=2\sin A\cos A

for every real AA, and

tan⁡2A=2tan⁡A1−tan⁡2A\tan2A=\frac{2\tan A}{1-\tan^2A}

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

In the formula booklet.

The tangent form requires the original tangent to exist. It also excludes tan⁡A=±1\tan A=\pm1, where cos⁡2A=0\cos2A=0 and the target tangent is undefined. If cos⁡A=0\cos A=0, this formula is unavailable even though tan⁡2A\tan2A can exist: for example, A=90∘A=90^\circ gives tan⁡180∘=0\tan180^\circ=0. Evaluate the target directly or use sine and cosine of 2A2A.

Worked example 3: doubling an acute angle

Question. Given sin⁡A=3/5\sin A=3/5 and 0∘<A<90∘0^\circ<A<90^\circ, find exact values of sin⁡2A\sin2A and tan⁡2A\tan2A.

Plan. Use the acute quadrant to choose the positive cosine, then form the double-angle sine and cosine.

cos⁡A=1−925=45positive in quadrant Isin⁡2A=2⋅35⋅45=2425cos⁡2A=1−2⋅925=725tan⁡2A=24/257/25=247.\begin{aligned} \cos A&=\sqrt{1-\frac9{25}}=\frac45 &&\text{positive in quadrant I}\\ \sin2A&=2\cdot\frac35\cdot\frac45=\frac{24}{25}\\ \cos2A&=1-2\cdot\frac9{25}=\frac7{25}\\ \tan2A&=\frac{24/25}{7/25}=\frac{24}{7}. \end{aligned}

Answer. sin⁡2A=24/25\sin2A=24/25 and tan⁡2A=24/7\tan2A=24/7.

Check. tan⁡A=3/4\tan A=3/4, so the tangent double-angle formula gives (3/2)/(1−9/16)=24/7(3/2)/(1-9/16)=24/7. Its denominator is nonzero.

Keep the given sine but move AA to quadrant II. Sine stays positive; cosine and tangent change sign.

Worked example 4: doubling an obtuse angle

Question. Given sin⁡A=3/5\sin A=3/5 and 90∘<A<180∘90^\circ<A<180^\circ, find exact values of sin⁡2A\sin2A and tan⁡2A\tan2A.

Plan. Choose the negative cosine before applying the double-angle identities.

cos⁡A=−45negative in quadrant IIsin⁡2A=2⋅35(−45)=−2425cos⁡2A=1−2⋅925=725tan⁡2A=−24/257/25=−247.\begin{aligned} \cos A&=-\frac45 &&\text{negative in quadrant II}\\ \sin2A&=2\cdot\frac35\left(-\frac45\right)=-\frac{24}{25}\\ \cos2A&=1-2\cdot\frac9{25}=\frac7{25}\\ \tan2A&=\frac{-24/25}{7/25}=-\frac{24}{7}. \end{aligned}

Answer. sin⁡2A=−24/25\sin2A=-24/25 and tan⁡2A=−24/7\tan2A=-24/7.

Check. Here A=143.130…∘A=143.130\ldots^\circ, so 2A=286.260…∘2A=286.260\ldots^\circ, in quadrant IV. Its sine and tangent are both negative, while its cosine is positive.

A ratio alone may not determine the sign of every doubled ratio. Always use any supplied angle interval before choosing a square-root sign.

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