P3 2.3 - Compound and double angles

P3 2.3 - Compound and double angles

Knowing the sine of two angles does not mean you can add their sine values to find the sine of their sum. P3 2.3 gives formulae for sums, differences and double angles; you will use them for exact values, half angles and algebraic proofs of identities.

Addition and subtraction formulae

Why is sin⁡(30∘+45∘)\sin(30^\circ+45^\circ) not sin⁡30∘+sin⁡45∘\sin30^\circ+\sin45^\circ? The second expression is greater than 11, so it cannot be a sine. Combining angles needs information about both horizontal and vertical coordinates.

Consider two unit-circle points (cos⁡A,sin⁡A)(\cos A,\sin A) and (cos⁡B,sin⁡B)(\cos B,\sin B). Their squared distance, found by subtracting coordinates, is

(cos⁡A−cos⁡B)2+(sin⁡A−sin⁡B)2=2−2(cos⁡Acos⁡B+sin⁡Asin⁡B).(\cos A-\cos B)^2+(\sin A-\sin B)^2 =2-2(\cos A\cos B+\sin A\sin B).

The cosine rule on the two unit radii gives the same squared chord length as 2−2cos⁡(A−B)2-2\cos(A-B). Equating the two gives the cosine difference formula. Replace BB by −B-B for the cosine sum formula, using cosine's even and sine's odd symmetry. Finally, applying the cosine difference formula to sin⁡(A+B)=cos⁡((π/2−A)−B)\sin(A+B)=\cos((\pi/2-A)-B) gives the sine sum formula; replacing BB by −B-B gives the sine difference.

Compound-angle formulae

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B,\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B, cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B.\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B.

Sine follows the sign of the angle operation; cosine uses the opposite sign in its second term. These formulae are printed in the P3 formula booklet.

Dividing the sine sum formula by the cosine sum formula, then dividing numerator and denominator by cos⁡Acos⁡B\cos A\cos B, gives

tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B.\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}.

Similarly,

tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B.\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}.

These are also supplied. Use the tangent versions where tan⁡A\tan A, tan⁡B\tan B and the resulting tangent are defined; a zero denominator signals an undefined tangent. The sine and cosine versions hold for all real angles in either unit, provided units are consistent.

Worked example 1

Without a calculator, find sin⁡75∘\sin75^\circ exactly.

Plan. Split 75∘75^\circ into 45∘+30∘45^\circ+30^\circ, whose sine and cosine values are known.

sin⁡75∘=sin⁡45∘cos⁡30∘+cos⁡45∘sin⁡30∘=2232+2212=6+24.\begin{aligned} \sin75^\circ &=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ\\ &=\frac{\sqrt2}{2}\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\frac12 =\frac{\sqrt6+\sqrt2}{4}. \end{aligned}

Answer. (6+2)/4(\sqrt6+\sqrt2)/4.

Check. This is approximately 0.9660.966, between sin⁡60∘\sin60^\circ and 11, as expected for an acute angle between 60∘60^\circ and 90∘90^\circ.

Keep the same angle split but change the requested function to tangent. The denominator now matters.

Worked example 2

Without a calculator, find tan⁡75∘\tan75^\circ exactly.

Plan. Apply the tangent addition formula with 45∘45^\circ and 30∘30^\circ, then rationalise.

tan⁡75∘=1+1/31−1/3=3+13−1=(3+1)23−1=2+3.\begin{aligned} \tan75^\circ &=\frac{1+1/\sqrt3}{1-1/\sqrt3} =\frac{\sqrt3+1}{\sqrt3-1}\\ &=\frac{(\sqrt3+1)^2}{3-1}=2+\sqrt3. \end{aligned}

Answer. 2+32+\sqrt3.

Check. Independently, 75∘=90∘−15∘75^\circ=90^\circ-15^\circ, so its tangent is the reciprocal of tan⁡15∘=(1−1/3)/(1+1/3)=2−3\tan15^\circ=(1-1/\sqrt3)/(1+1/\sqrt3)=2-\sqrt3. Their product is (2+3)(2−3)=1(2+\sqrt3)(2-\sqrt3)=1.

Keep the operations distinct when choosing the formula.

Watch out

sin⁡(A+B)\sin(A+B) is not sin⁡A+sin⁡B\sin A+\sin B. Each term in the correct formula contains one sine and one cosine. In the tangent sum formula, the denominator is 1−tan⁡Atan⁡B1-\tan A\tan B; its sign is opposite to the numerator's operation.

Choosing a useful angle split makes exact work possible. The formula does not require AA and BB themselves to be acute; retain their signed sine and cosine values.

Double-angle identities

A double angle is a compound angle with the same angle used twice. Put A=B=θA=B=\theta into the addition formulae:

sin⁡(θ+θ)=sin⁡θcos⁡θ+cos⁡θsin⁡θ=2sin⁡θcos⁡θ,\sin(\theta+\theta)=\sin\theta\cos\theta+\cos\theta\sin\theta =2\sin\theta\cos\theta, cos⁡(θ+θ)=cos⁡2θ−sin⁡2θ.\cos(\theta+\theta)=\cos^2\theta-\sin^2\theta.

Using sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 gives two further cosine forms. Choose the one using the function you already know.

Double-angle identities

sin⁡2θ=2sin⁡θcos⁡θ,\sin2\theta=2\sin\theta\cos\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, tan⁡2θ=2tan⁡θ1−tan⁡2θ.\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}.

The tangent result follows by putting A=B=θA=B=\theta in its sum formula, where all terms are defined. These double-angle identities are expected recall, not printed as double-angle results in the P3 booklet.

For instance, if tan⁡θ=2\tan\theta=2, then tan⁡2θ=4/(1−4)=−4/3\tan2\theta=4/(1-4)=-4/3. The sign can change because doubling the angle can move it into a different quadrant. Also sin⁡2θ\sin2\theta and 2sin⁡θ2\sin\theta are different expressions: the first doubles the angle before evaluating sine.

Worked example 3

Given sin⁡θ=3/5\sin\theta=3/5 and 0<θ<π/20<\theta<\pi/2, find sin⁡2θ\sin2\theta exactly.

Plan. Obtain positive cosine from Pythagoras, then use the sine double-angle identity.

cos⁡θ=1−925=45,sin⁡2θ=23545=2425.\cos\theta=\sqrt{1-\frac9{25}}=\frac45, \qquad \sin2\theta=2\frac35\frac45=\frac{24}{25}.

Answer. 24/2524/25.

Check. The original angle is approximately 36.87∘36.87^\circ, so the double angle is about 73.74∘73.74^\circ. Its sine is close to, but less than, 11, consistent with 24/25=0.9624/25=0.96.

Keep the given sine and interval but change the requested function to cosine. A form using sine alone avoids an unnecessary square root.

Worked example 4

Given sin⁡θ=3/5\sin\theta=3/5 and 0<θ<π/20<\theta<\pi/2, find cos⁡2θ\cos2\theta exactly.

Plan. Use 1−2sin⁡2θ1-2\sin^2\theta.

cos⁡2θ=1−2(35)2=1−1825=725.\cos2\theta=1-2\left(\frac35\right)^2 =1-\frac{18}{25}=\frac7{25}.

Answer. 7/257/25.

Check. Using the two-square form independently gives (4/5)2−(3/5)2=7/25(4/5)^2-(3/5)^2=7/25. Together with the previous answer, (7/25)2+(24/25)2=1(7/25)^2+(24/25)^2=1.

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