1.05l, 1.05m - Compound and double angle formulae

1.05l, 1.05m - Compound and double angle formulae

Build angle-addition formulae from a geometric picture, derive double-angle identities, and choose the right form for exact values, simplification and bounded equations.

A geometric proof of angle addition

How can familiar angles give the exact value of sin⁡15∘\sin15^\circ? Start with the easier question of what happens when a unit direction turns through two angles, one after the other. Turning through 30∘30^\circ, then 45∘45^\circ, must reach the same point as turning through 75∘75^\circ.

You already know

A unit direction at angle AA has components (cos⁡A,sin⁡A)(\cos A,\sin A). A right-angle anticlockwise turn sends (x,y)(x,y) to (−y,x)(-y,x). Components add even when they are negative.

Take those perpendicular directions as the starting point. A further turn BB has signed components cos⁡B\cos B along the first direction and sin⁡B\sin B along the perpendicular one.

A unit direction is resolved along two perpendicular directions before its horizontal and vertical components are added

The picture shows acute angles so the projection triangle is easy to see. Signed components make the same argument work in every quadrant.

Compound-angle identity

A compound-angle identity expresses a trigonometric function of a sum or difference using functions of the individual angles. It holds for every input in its stated domain; an equation such as sin⁡x=1/2\sin x=1/2 instead holds only at particular inputs.

Read each coordinate separately to see why these identities have different signs.

Worked example 1: recover the vertical component

Question. Use the perpendicular unit directions (cos⁡A,sin⁡A)(\cos A,\sin A) and (−sin⁡A,cos⁡A)(-\sin A,\cos A) to establish the sine-addition identity.

Plan. Add the vertical contributions to the final unit direction.

vertical component=(cos⁡B)sin⁡A+(sin⁡B)cos⁡Aadd projectionssin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡Bidentify final angle\begin{aligned} \text{vertical component}&=(\cos B)\sin A+(\sin B)\cos A &&\text{add projections}\\ \sin(A+B)&=\sin A\cos B+\cos A\sin B &&\text{identify final angle} \end{aligned}

Why this step? The final direction is built from two perpendicular components; sine reads off their combined vertical component.

Answer. sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B)=\sin A\cos B+\cos A\sin B, for real A,BA,B.

Check. At A=B=30∘A=B=30^\circ, the right side is 2(1/2)(3/2)=3/2=sin⁡60∘2(1/2)(\sqrt3/2)=\sqrt3/2=\sin60^\circ.

Change only the component being read: the horizontal contribution from the perpendicular direction has a minus sign.

Worked example 2: recover the horizontal component

Question. Establish the cosine-addition identity from the same directions.

Plan. Add horizontal contributions rather than vertical ones.

horizontal component=(cos⁡B)cos⁡A+(sin⁡B)(−sin⁡A)signed projectionscos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡Bidentify final angle\begin{aligned} \text{horizontal component}&=(\cos B)\cos A+(\sin B)(-\sin A) &&\text{signed projections}\\ \cos(A+B)&=\cos A\cos B-\sin A\sin B &&\text{identify final angle} \end{aligned}

Why this step? The perpendicular direction points left when AA is acute, which explains the subtraction.

Answer. cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B)=\cos A\cos B-\sin A\sin B, for real A,BA,B.

Check. At A=B=45∘A=B=45^\circ, the right side is 1/2−1/2=0=cos⁡90∘1/2-1/2=0=\cos90^\circ.

Here are the two results together. The geometric argument proves them; quoting them would not prove their geometric origin.

Sine and cosine addition

For all real angles A,BA,B, in the same units,

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B,cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B.\sin(A+B)=\sin A\cos B+\cos A\sin B, \qquad \cos(A+B)=\cos A\cos B-\sin A\sin B.

In the formula sheet.

Sine does not distribute over addition: sin⁡(90∘+90∘)=0\sin(90^\circ+90^\circ)=0, whereas sin⁡90∘+sin⁡90∘=2\sin90^\circ+\sin90^\circ=2. The missing product terms make a spectacular difference.

Difference formulae and tangent domains

A subtraction formula is an addition formula with a negative second angle. Use sin⁡(−B)=−sin⁡B\sin(-B)=-\sin B and cos⁡(−B)=cos⁡B\cos(-B)=\cos B.

Difference and tangent identities

For all real A,BA,B,

sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B,cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B.\sin(A-B)=\sin A\cos B-\cos A\sin B, \qquad \cos(A-B)=\cos A\cos B+\sin A\sin B.

Also,

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},

provided the individual tangents exist and the displayed denominator is nonzero.

In the formula sheet.

To see the tangent restriction, divide the sine-addition identity by the cosine-addition identity, then divide both numerator and denominator by cos⁡Acos⁡B\cos A\cos B. That last division requires both factors to be nonzero; the first requires cos⁡(A+B)≠0\cos(A+B)\ne0.

Worked example 3: a tangent sum

Question. Given tan⁡A=1/2\tan A=1/2 and tan⁡B=1/3\tan B=1/3, find tan⁡(A+B)\tan(A+B).

Plan. Substitute into the sum identity and inspect its denominator.

tan⁡(A+B)=1/2+1/31−(1/2)(1/3)sum identity=5/65/6=1nonzero denominator\begin{aligned} \tan(A+B)&=\frac{1/2+1/3}{1-(1/2)(1/3)} &&\text{sum identity}\\ &=\frac{5/6}{5/6}=1 &&\text{nonzero denominator} \end{aligned}

Why this step? The sign in the denominator is opposite to the sign of the angle combination.

Answer. tan⁡(A+B)=1\tan(A+B)=1.

Check. Choosing the acute representatives gives angles about 26.565∘26.565^\circ and 18.435∘18.435^\circ, whose sum is 45∘45^\circ.

Change only the angle combination from a sum to a difference; both signs must change together.

Worked example 4: a tangent difference

Question. With tan⁡A=1/2\tan A=1/2 and tan⁡B=1/3\tan B=1/3, find tan⁡(A−B)\tan(A-B).

Plan. Use the difference identity with a plus in its denominator.

tan⁡(A−B)=1/2−1/31+(1/2)(1/3)difference identity=1/67/6=17simplify fractions\begin{aligned} \tan(A-B)&=\frac{1/2-1/3}{1+(1/2)(1/3)} &&\text{difference identity}\\ &=\frac{1/6}{7/6}=\frac17 &&\text{simplify fractions} \end{aligned}

Why this step? The denominator is 1+tan⁡Atan⁡B1+\tan A\tan B; copying the sum denominator would change the result.

Answer. tan⁡(A−B)=1/7\tan(A-B)=1/7.

Check. For acute representatives, the difference is positive and small, as is 1/71/7.

An undefined representation does not always mean an undefined target.

Watch out

At A=π/2,B=π/4A=\pi/2,B=\pi/4, tan⁡(A+B)=−1\tan(A+B)=-1, but the tangent-addition fraction cannot be used because tan⁡A\tan A is undefined. Use sine and cosine addition instead. At A=B=π/4A=B=\pi/4, the final tangent really is undefined: the denominator is zero.

Keep that domain check even when the arithmetic looks straightforward.

Exact values from familiar angles

The opening question now has a route: 15∘=45∘−30∘15^\circ=45^\circ-30^\circ. Exact values for familiar angles can manufacture exact values for unfamiliar ones.

You already know

sin⁡30∘=1/2\sin30^\circ=1/2, cos⁡30∘=3/2\cos30^\circ=\sqrt3/2, and sin⁡45∘=cos⁡45∘=2/2\sin45^\circ=\cos45^\circ=\sqrt2/2. Exact surds stay exact throughout a calculation.

Choose the angle decomposition before choosing the identity.

Continue this lesson.

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