Mt16, Mt17 - Compound and double angle formulae | OCR A-Level Mathematics B (MEI)

Use a supplied geometric starting point to justify angle identities, derive double-angle results, and read a short argument that turns a multiple angle into an algebraic ratio.

A geometric proof of angle addition

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

You already know

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

Here the perpendicular directions are the supplied starting point. Read what they mean before starting the proof. A further turn $B$ has signed components $\cos B$ along the first direction and $\sin B$ 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 $\sin x=1/2$ 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 $(\cos A,\sin A)$ and $(-\sin A,\cos A)$ to establish the sine-addition identity.

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

\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. $\sin(A+B)=\sin A\cos B+\cos A\sin B$ , for real $A,B$ .

Check. At $A=B=30^\circ$ , the right side is $2(1/2)(\sqrt3/2)=\sqrt3/2=\sin60^\circ$ .

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.

\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 $A$ is acute, which explains the subtraction.

Answer. $\cos(A+B)=\cos A\cos B-\sin A\sin B$ , for real $A,B$ .

Check. At $A=B=45^\circ$ , the right side is $1/2-1/2=0=\cos90^\circ$ .

Here are the two results together. The projection argument proves them from the supplied starting point; quoting the identities would not explain why they follow.

Sine and cosine addition

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

\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: $\sin(90^\circ+90^\circ)=0$ , whereas $\sin90^\circ+\sin90^\circ=2$ . 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 $\sin(-B)=-\sin B$ and $\cos(-B)=\cos B$ .

Difference and tangent identities

For all real $A,B$ ,

\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\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 $\cos A\cos B$ . That last division requires both factors to be nonzero; the first requires $\cos(A+B)\ne0$ .

Worked example 3: a tangent sum

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

Plan. Substitute into the sum identity and inspect its 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. $\tan(A+B)=1$ .

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

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

Worked example 4: a tangent difference

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

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

\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 $1+\tan A\tan B$ ; copying the sum denominator would change the result.

Answer. $\tan(A-B)=1/7$ .

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

An undefined representation does not always mean an undefined target.

Watch out

At $A=\pi/2,B=\pi/4$ , $\tan(A+B)=-1$ , but the tangent-addition fraction cannot be used because $\tan A$ is undefined. Use sine and cosine addition instead. At $A=B=\pi/4$ , 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

Our exact-value question now has a route: $15^\circ=45^\circ-30^\circ$ . Exact values for familiar angles can manufacture exact values for unfamiliar ones.

You already know

$\sin30^\circ=1/2$ , $\cos30^\circ=\sqrt3/2$ , and $\sin45^\circ=\cos45^\circ=\sqrt2/2$ . Exact surds stay exact throughout a calculation.

Choose the angle decomposition before choosing the identity.

Worked example 5: exact cosine of a difference

Question. Determine $\cos15^\circ$ exactly.

Plan. Write $15^\circ=45^\circ-30^\circ$ and use cosine subtraction.

\begin{aligned} \cos15^\circ&=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ &&\text{subtract angles}\\ &=\frac{\sqrt2}{2}\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\frac12 &&\text{exact values}\\ &=\frac{\sqrt6+\sqrt2}{4} &&\text{collect surds} \end{aligned}

Why this step? Both component angles have known exact values, so no inverse function or decimal approximation is needed.

Answer. $\cos15^\circ=(\sqrt6+\sqrt2)/4$ .

Check. This is about $0.966$ , between $\cos30^\circ=\sqrt3/2$ and $1$ .

Change only the sign joining those same component angles.

Worked example 6: exact cosine of a sum

Question. Determine $\cos75^\circ$ exactly.

Plan. Use $75^\circ=45^\circ+30^\circ$ .

\begin{aligned} \cos75^\circ&=\cos45^\circ\cos30^\circ-\sin45^\circ\sin30^\circ &&\text{add angles}\\ &=\frac{\sqrt6}{4}-\frac{\sqrt2}{4} &&\text{exact products}\\ &=\frac{\sqrt6-\sqrt2}{4} &&\text{combine} \end{aligned}

Why this step? The same products occur; only their connecting sign changes.

Answer. $\cos75^\circ=(\sqrt6-\sqrt2)/4$ .

Check. The result is about $0.259$ , positive and below $1/2$ .

The opening sine calculation uses subtraction too: $\sin15^\circ=(\sqrt6-\sqrt2)/4$ . Also $\tan75^\circ=(1+1/\sqrt3)/(1-1/\sqrt3)=2+\sqrt3$ , after rationalising. In radians, $15^\circ=\pi/12$ and $75^\circ=5\pi/12$ : the identities have not changed.

Exam tip

For “Show that”, display the intermediate identity and substitution. Copying the supplied result, or checking one angle on a calculator, does not establish it for every allowed input.

Try the companion cosine calculation, whose sign tells you which quadrant contains the angle.

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Mt16, Mt17 - Compound and double angle formulae

A geometric proof of angle addition

Difference formulae and tangent domains

Exact values from familiar angles

Continue this lesson.

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