E6 - Compound and double-angle formulae

E6 - Compound and double-angle formulae

Derive compound-angle formulae from geometric projections and use them to find exact values. Obtain the double-angle formulae from those results, choose a form that fits the information, and connect the formulae in proofs.

A geometric proof of addition formulae

Can you find sin⁡75∘\sin75^\circ exactly using only familiar right triangles? Since 75∘=45∘+30∘75^\circ=45^\circ+30^\circ, the question asks how the height of a unit length changes when two rotations are combined. It is not the sum of two separate sine values: sin⁡45∘+sin⁡30∘>1\sin45^\circ+\sin30^\circ>1, which no sine can be.

You already know

A length ll at angle AA to the horizontal has horizontal and vertical projections lcos⁡Al\cos A and lsin⁡Al\sin A. For the unit circle these become cosine and sine themselves.

Start with acute A,BA,B and A+B<90∘A+B<90^\circ. The right triangle below has OP=1OP=1, right angle OQPOQP, and angle QOP=BQOP=B. The ray OQOQ makes angle AA with the horizontal.

Unit right triangle with angles A and B, showing the horizontal and vertical projections used to prove both addition formulae

The first right triangle gives OQ=cos⁡BOQ=\cos B and QP=sin⁡BQP=\sin B. Project the first length:

OK=cos⁡Bcos⁡AQK=cos⁡Bsin⁡AOK=\cos B\cos A\qquad QK=\cos B\sin A

Because QPQP is perpendicular to OQOQ, it makes angle AA with the vertical. Therefore its upward and leftward projections are

PN=sin⁡Bcos⁡AQN=sin⁡Bsin⁡APN=\sin B\cos A\qquad QN=\sin B\sin A

The height of PP adds the upward contributions. Its horizontal coordinate subtracts the leftward contribution:

PH=QK+PN=sin⁡Acos⁡B+cos⁡Asin⁡BOH=OK−QN=cos⁡Acos⁡B−sin⁡Asin⁡B\begin{aligned} PH&=QK+PN=\sin A\cos B+\cos A\sin B\\ OH&=OK-QN=\cos A\cos B-\sin A\sin B \end{aligned}

But the unit length OPOP makes angle A+BA+B with the horizontal. Its height is sin⁡(A+B)\sin(A+B), and its horizontal coordinate is cos⁡(A+B)\cos(A+B). This proves both formulae in the pictured configuration.

For other angles, use directed projections: right/up are positive, left/down negative. A unit direction at AA has components (cos⁡A,sin⁡A)(\cos A,\sin A); its anticlockwise perpendicular has components (−sin⁡A,cos⁡A)(-\sin A,\cos A), by right-angle symmetry of the unit circle. The geometric decomposition of a unit length at angle BB from that direction has coefficients cos⁡B\cos B and sin⁡B\sin B. Multiply the two component pairs by those coefficients and add. The same expressions follow for any oriented angles; negative coefficients reverse direction. This extends the geometric argument without assuming an addition formula.

Sine and cosine addition

For all real angles, using one consistent angle unit,

sin⁡(A+B)≡sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B)\equiv\sin A\cos B+\cos A\sin B cos⁡(A+B)≡cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B)\equiv\cos A\cos B-\sin A\sin B

In the formula booklet.

Now answer the opening puzzle directly from the height formula:

sin⁡75∘=2232+2212=6+24\sin75^\circ=\frac{\sqrt2}{2}\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\frac12 =\frac{\sqrt6+\sqrt2}{4}

There is a small surprise here: the square roots of 22 and 33, coming from two different triangles, combine into an exact value at a new angle. No new triangle table is needed.

Key idea

The sine formula adds two vertical projections; the cosine formula subtracts a leftward projection. Their different signs come from the geometry.

Use the picture to explain those signs, rather than reciting the finished formulae.

Difference and tangent formulae

A negative rotation supplies the difference formulae. Since sin⁡(−B)=−sin⁡B\sin(-B)=-\sin B and cos⁡(−B)=cos⁡B\cos(-B)=\cos B, replacing BB by −B-B reverses only the sine contributions.

The even/odd and right-angle symmetries used here follow from the unit circle; they are not separately printed in the formula booklet. For example, a reflection in the horizontal axis preserves the horizontal coordinate and reverses the vertical one.

Difference formulae

For all real angles in a consistent unit,

sin⁡(A−B)≡sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A-B)\equiv\sin A\cos B-\cos A\sin B cos⁡(A−B)≡cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A-B)\equiv\cos A\cos B+\sin A\sin B

In the formula booklet.

Tangent is the ratio of the vertical to the horizontal projection in the same construction. Taking that ratio gives

tan⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡Bcos⁡Acos⁡B−sin⁡Asin⁡B=tan⁡A+tan⁡B1−tan⁡Atan⁡B\begin{aligned} \tan(A+B)&=\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B-\sin A\sin B}\\ &=\frac{\tan A+\tan B}{1-\tan A\tan B} \end{aligned}

The first denominator is nonzero where tan⁡(A+B)\tan(A+B) exists. The second line additionally divides top and bottom by cos⁡Acos⁡B\cos A\cos B, so both individual tangents must exist. Substituting −B-B also gives the tangent difference formula.

Tangent of a sum or difference

tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}

This form requires cos⁡A≠0\cos A\ne0, cos⁡B≠0\cos B\ne0 and cos⁡(A±B)≠0\cos(A\pm B)\ne0. Its denominator cannot be zero.

In the formula booklet.

Choose familiar component angles before substituting. Exact values keep the result exact.

Worked example 1: tangent of a sum

Question. Use a compound-angle formula to find tan⁡75∘\tan75^\circ exactly.

Plan. Write 75∘=45∘+30∘75^\circ=45^\circ+30^\circ, substitute exact tangents, and rationalise.

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

Why this step? Rationalising turns the exact quotient into a simple surd while the denominator remains nonzero.

Answer. tan⁡75∘=2+3\tan75^\circ=2+\sqrt3.

Check. 75∘75^\circ is acute and exceeds 45∘45^\circ, so a positive tangent greater than 11 is sensible.

Change only addition to subtraction in the component angles. The denominator's sign changes with it.

Worked example 2: tangent of a difference

Question. Use a compound-angle formula to find tan⁡15∘\tan15^\circ exactly.

Plan. Use 45∘−30∘45^\circ-30^\circ with the plus sign in the denominator.

tan⁡15∘=1−1/31+1/3difference formula=3−13+1clear the small fractions=(3−1)23−1=2−3rationalise\begin{aligned} \tan15^\circ&=\frac{1-1/\sqrt3}{1+1/\sqrt3} &&\text{difference formula}\\ &=\frac{\sqrt3-1}{\sqrt3+1} &&\text{clear the small fractions}\\ &=\frac{(\sqrt3-1)^2}{3-1}=2-\sqrt3 &&\text{rationalise} \end{aligned}

Why this step? For tangent the numerator follows the angle sign and the denominator takes the opposite sign.

Answer. tan⁡15∘=2−3\tan15^\circ=2-\sqrt3.

Check. The product (2+3)(2−3)=1(2+\sqrt3)(2-\sqrt3)=1, agreeing with the complementary angles' reciprocal tangents.

The formula expresses a ratio, not a distributive law for tangent.

Watch out

Wrong: tan⁡(A−B)=tan⁡A−tan⁡B\tan(A-B)=\tan A-\tan B. This omits the denominator. Right: divide that difference by 1+tan⁡Atan⁡B1+\tan A\tan B, checking that the terms exist.

Keep the difference form but change the second component angle to 60∘60^\circ. A negative angle now provides a useful sign check.

Continue this lesson.

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