Trigonometric Functions of Sum and Difference of Angles

Compound angles

A compound angle is the sum or difference of two (or more) angles, such as $A+B$ or $A-B$. This section

proves how the six trigonometric ratios of a compound angle relate to the ratios of the two original angles.

Theorem 1: $\cos(A-B)=\cos A\cos B+\sin A\sin B$

Proof (unit circle method). Draw a unit circle centred at the origin $O$. Mark point $P$ on the circle so

that $OP$ makes angle $A$ with the positive $x$-axis, and point $Q$ so that $OQ$ makes angle $B$. Since the

circle has radius $1$, the coordinates are $P\equiv(\cos A,\sin A)$ and $Q\equiv(\cos B,\sin B)$.

Using the distance formula in the original coordinate system:

Expanding: $=\cos^2A-2\cos A\cos B+\cos^2B+\sin^2A-2\sin A\sin B+\sin^2B=2-2(\cos A\cos B+\sin A\sin B)$

(using $\sin^2+\cos^2=1$ twice). Call this result (1).

Now imagine rotating the axis so $OQ$ becomes the new positive $x$-axis. In this new system, $\angle QOP=A-B$,

so $P\equiv(\cos(A-B),\sin(A-B))$ and $Q\equiv(1,0)$. The distance $PQ$ is unchanged by a rotation, so:

Call this (2).

Equating (1) and (2): $2-2\cos(A-B)=2-2(\cos A\cos B+\sin A\sin B)$, which gives

Theorem 2: $\cos(A+B)=\cos A\cos B-\sin A\sin B$

Proof. Put $x=A,\ y=-B$ into Theorem 1's formula $\cos(x-y)=\cos x\cos y+\sin x\sin y$:

Since $\cos(-\theta)=\cos\theta$ and $\sin(-\theta)=-\sin\theta$: $\cos(A+B)=\cos A\cos B-\sin A\sin B$.

Results at $B=\pi/2$ (a first glimpse of allied angles)

Putting $x=\pi/2,y=\theta$ into $\cos(x-y)=\cos x\cos y+\sin x\sin y$ gives $\cos(\pi/2-\theta)=0\cdot\cos\theta+ 1\cdot\sin\theta=\sin\theta$. Similarly: $\cos(\pi/2+\theta)=-\sin\theta$, $\sin(\pi/2-\theta)=\cos\theta$,

$\sin(\pi/2+\theta)=\cos\theta$, $\tan(\pi/2-\theta)=\cot\theta$, $\tan(\pi/2+\theta)=-\cot\theta$. These become

the seed of the full allied-angle table in section 3.2.

Theorem 3: $\sin(A-B)=\sin A\cos B-\cos A\sin B$

Proof. Using $\cos(\pi/2-\theta)=\sin\theta$ with $\theta=A-B$: $\sin(A-B)=\cos[\pi/2-(A-B)]=\cos[(\pi/2-A)+B]$.

Expand with the cosine-sum formula: $=\cos(\pi/2-A)\cos B-\sin(\pi/2-A)\sin B=\sin A\cos B-\cos A\sin B$.

Theorem 4: $\sin(A+B)=\sin A\cos B+\cos A\sin B$

Proved the same way, replacing $B$ by $-B$ in Theorem 3.

Theorem 5: $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$

Proof. $\tan(A+B)=\dfrac{\sin(A+B)}{\cos(A+B)}=\dfrac{\sin A\cos B+\cos A\sin B}{\cos A\cos B-\sin A\sin B}$.

Divide numerator and denominator by $\cos A\cos B$:

Theorem 6: $\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$

Proved the same way with $\cos(A-B),\sin(A-B)$ (an Activity in the textbook — same divide-by-$\cos A\cos B$

technique as Theorem 5, with $B$ replaced by $-B$).

Results (cotangent versions, provided none of $A,B,A\pm B$ is a multiple of $\pi$):

(taking reciprocals of Theorems 5 and 6).

Solved Examples

Ex.1 — Find $\cos15°$. Write $15°=45°-30°$. By Theorem 1: $\cos15°=\cos45°\cos30°+\sin45°\sin30° =\dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt3}{2}+\dfrac{1}{\sqrt2}\cdot\dfrac12=\dfrac{\sqrt3+1}{2\sqrt2}$.

Ex.2 — Find $\tan(13\pi/12)$. Write $\dfrac{13\pi}{12}=\pi+\dfrac{\pi}{12}$; since $\tan(\pi+\theta)=\tan\theta$,

this is $\tan\dfrac{\pi}{12}$. Write $\dfrac{\pi}{12}=\dfrac{\pi}{4}-\dfrac{\pi}{6}$ and apply Theorem 6:

$\tan\dfrac{\pi}{12}=\dfrac{\tan\frac{\pi}{4}-\tan\frac{\pi}{6}}{1+\tan\frac{\pi}{4}\tan\frac{\pi}{6}} =\dfrac{1-\frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}}=2-\sqrt3$.

Ex.3 — Show $\dfrac{\sin(x+y)}{\sin(x-y)}=\dfrac{\tan x+\tan y}{\tan x-\tan y}$. Expand numerator and denominator

with Theorems 3–4, then divide every term by $\cos x\cos y$ to convert to tangents; both sides match.

Ex.4 — Show $\tan3x\tan2x\tan x=\tan3x-\tan2x-\tan x$. Write $\tan3x=\tan(2x+x)$ via Theorem 5, cross-multiply

$\tan3x(1-\tan2x\tan x)=\tan2x+\tan x$, and rearrange to $\tan3x-\tan2x-\tan x=\tan3x\tan2x\tan x$.

Ex.5 — Show $\cos(x+\pi/4)+\cos(x-\pi/4)=\sqrt2\cos x$. Expand each cosine with $\cos(\pi/4)=\sin(\pi/4) =1/\sqrt2$: each gives $\frac{1}{\sqrt2}\cos x\mp\frac{1}{\sqrt2}\sin x$; the two $\sin x$ terms cancel, leaving

$\frac{2}{\sqrt2}\cos x=\sqrt2\cos x$.

Ex.6 — If $\tan A-\tan B=x$ and $\cot B-\cot A=y$, show $\cot(A-B)=\dfrac{1}{\frac1x+\frac1y}$. From

$\cot B-\cot A=y$: $\dfrac{\tan A-\tan B}{\tan A\tan B}=y\Rightarrow\dfrac{x}{\tan A\tan B}=y\Rightarrow\tan A\tan B =\dfrac xy$. Then $\cot(A-B)=\dfrac{1+\tan A\tan B}{\tan A-\tan B}=\dfrac{1+x/y}{x}=\dfrac{x+y}{xy}=\dfrac1x+\dfrac1y =\dfrac{1}{\frac1x+\frac1y}$ (written as a single reciprocal-sum).

Ex.7 — If $\tan\alpha=\dfrac{1}{x+1}$, $\tan\beta=\dfrac{1}{2x+1}$, $\tan\gamma=\dfrac{x+1}{x^2+x+1}$, show $\alpha+\beta=\gamma$. Compute $\tan(\alpha+\beta)$ using Theorem 5; after clearing denominators the fraction

simplifies (via a common-denominator combination and cancellation) to $\dfrac{x+1}{x^2+x+1}$, which is exactly

$\tan\gamma$.

Ex.8 — If $\sin A+\sin B=x$, $\cos A+\cos B=y$, show $\sin(A+B)=\dfrac{2xy}{x^2+y^2}$. Compute $x^2+y^2=2+2\cos(A-B)$

and $y^2-x^2=\cos(A+B)(x^2+y^2)$ (both by expanding the squares and using Theorems 1–2), so

$\cos(A+B)=\dfrac{y^2-x^2}{x^2+y^2}$. Then $\sin(A+B)=\sqrt{1-\cos^2(A+B)}$ simplifies, after combining over a

common denominator and factoring, to $\dfrac{2xy}{x^2+y^2}$.