Sum and difference formulas
If you know the sines and cosines of two angles ${A}$ and ${B}$, can you find ${\cos(A+B)}$? Yes, but not by adding. This page gives the correct formulas, shows why they are true, and uses them to find exact values and to simplify expressions.
Watch out: ${\cos(A+B)}$ is not ${\cos A+\cos B}$. For example, ${\cos(30^\circ+60^\circ)=\cos 90^\circ=0}$. But ${\cos 30^\circ+\cos 60^\circ}$ is ${\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}}$, which is not ${0}$.
The formulas
$\begin{align*}&\sin(A+B)\\&=\sin A\cos B+\cos A\sin B\end{align*}$
$\begin{align*}&\sin(A-B)\\&=\sin A\cos B-\cos A\sin B\end{align*}$
$\begin{align*}&\cos(A+B)\\&=\cos A\cos B-\sin A\sin B\end{align*}$
$\begin{align*}&\cos(A-B)\\&=\cos A\cos B+\sin A\sin B\end{align*}$
$\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
$\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$
Some patterns help you remember them:
- The sine formulas mix: sine-cosine, then cosine-sine. They keep the sign: ${A+B}$ gives ${+}$.
- The cosine formulas pair like with like: cosine-cosine, then sine-sine. They switch the sign: ${A+B}$ gives ${-}$.
Why the formulas are true
We prove the formula for ${\cos(A-B)}$ first. The others follow from it.
Put the angles ${A}$ and ${B}$ in standard position. Their terminal points on the unit circle are ${P(\cos A,\sin A)}$ and ${Q(\cos B,\sin B)}$. The angle between them is ${A-B}$, as in figure (a). Now turn the whole picture until ${Q}$ lands on ${S(1,0)}$, as in figure (b). Then ${P}$ lands on ${R(\cos(A-B),\sin(A-B))}$.
Turning does not change lengths, so ${RS=PQ}$. Find both with the distance formula, and use ${\sin^2+\cos^2=1}$ to simplify.
Length of ${PQ}$:
$\begin{align*}PQ^2&=(\cos A-\cos B)^2\\&\quad+(\sin A-\sin B)^2\\&=2-2(\cos A\cos B+\sin A\sin B)\end{align*}$
Length of ${RS}$:
$\begin{align*}RS^2&=(\cos(A-B)-1)^2\\&\quad+\sin^2(A-B)\\&=2-2\cos(A-B)\end{align*}$
Set the two equal. The ${2}$'s cancel, and dividing by ${-2}$ gives
$\begin{align*}&\cos(A-B)\\&=\cos A\cos B+\sin A\sin B\end{align*}$
The other formulas. Replace ${B}$ by ${-B}$. Since ${\cos(-B)=\cos B}$ and ${\sin(-B)=-\sin B}$, this gives the formula for ${\cos(A+B)}$. The sine formulas come from the cofunction identity ${\sin x=\cos\left(\dfrac{\pi}{2}-x\right)}$. The tangent formulas come from dividing the sine formula by the cosine formula.
Exact values
Many angles can be written as a sum or difference of ${30^\circ}$, ${45^\circ}$, ${60^\circ}$, and their multiples. Then the formulas give exact values.
Example 1: Find the exact value of ${\cos 75^\circ}$.
Solution:
Write ${75^\circ=45^\circ+30^\circ}$, and use the formula for ${\cos(A+B)}$:
$\begin{align*}&\cos 75^\circ\\&=\cos 45^\circ\cos 30^\circ-\sin 45^\circ\sin 30^\circ\\&=\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}\\&=\dfrac{\sqrt{6}-\sqrt{2}}{4}\end{align*}$
Check: $\dfrac{\sqrt{6}-\sqrt{2}}{4}\approx 0.2588$, and a calculator gives ${\cos 75^\circ\approx 0.2588}$.
Example 2: Find the exact value of ${\sin\dfrac{\pi}{12}}$.
Solution:
Write $\dfrac{\pi}{12}=\dfrac{\pi}{3}-\dfrac{\pi}{4}$ (that is, ${15^\circ=60^\circ-45^\circ}$). Use the formula for ${\sin(A-B)}$:
$\begin{align*}&\sin\dfrac{\pi}{12}\\&=\sin\dfrac{\pi}{3}\cos\dfrac{\pi}{4}-\cos\dfrac{\pi}{3}\sin\dfrac{\pi}{4}\\&=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}\\&=\dfrac{\sqrt{6}-\sqrt{2}}{4}\end{align*}$
This equals ${\cos 75^\circ}$ from Example 1. That makes sense: ${15^\circ}$ and ${75^\circ}$ are complementary.
Example 3: Find the exact value of ${\tan 75^\circ}$.
Solution:
Use ${75^\circ=45^\circ+30^\circ}$, with ${\tan 45^\circ=1}$ and ${\tan 30^\circ=\dfrac{1}{\sqrt{3}}}$:
$\tan 75^\circ=\dfrac{1+\dfrac{1}{\sqrt{3}}}{1-\dfrac{1}{\sqrt{3}}}$
Multiply the top and bottom by ${\sqrt{3}}$, then by the conjugate ${\sqrt{3}+1}$:
$\begin{align*}&\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{3}+1}\\&=\dfrac{3+2\sqrt{3}+1}{3-1}\\&=\dfrac{4+2\sqrt{3}}{2}=2+\sqrt{3}\end{align*}$
Example 4: Find the exact value of $\sin 20^\circ\cos 40^\circ+\cos 20^\circ\sin 40^\circ$.
Solution:
This is the right side of the formula for ${\sin(A+B)}$, with ${A=20^\circ}$ and ${B=40^\circ}$. So it equals
$\sin(20^\circ+40^\circ)=\sin 60^\circ=\dfrac{\sqrt{3}}{2}$
Using the formulas with given values
Example 5: Suppose ${\sin A=\dfrac{3}{5}}$ with ${A}$ in quadrant I, and ${\cos B=-\dfrac{5}{13}}$ with ${B}$ in quadrant II. Find ${\sin(A+B)}$.
Solution:
We need ${\cos A}$ and ${\sin B}$. Use ${\sin^2+\cos^2=1}$ and the quadrants:
$\cos A=\sqrt{1-\dfrac{9}{25}}=\dfrac{4}{5}$ (positive in quadrant I)
$\sin B=\sqrt{1-\dfrac{25}{169}}=\dfrac{12}{13}$ (positive in quadrant II)
Now use the formula:
$\begin{align*}&\sin(A+B)\\&=\sin A\cos B+\cos A\sin B\\&=\dfrac{3}{5}\cdot\left(-\dfrac{5}{13}\right)+\dfrac{4}{5}\cdot\dfrac{12}{13}\\&=\dfrac{-15+48}{65}=\dfrac{33}{65}\end{align*}$
Example 6: Prove the cofunction identity ${\cos\left(\dfrac{\pi}{2}-x\right)=\sin x}$.
Solution:
Use the formula for ${\cos(A-B)}$, with ${\cos\dfrac{\pi}{2}=0}$ and ${\sin\dfrac{\pi}{2}=1}$:
$\begin{align*}&\cos\left(\dfrac{\pi}{2}-x\right)\\&=\cos\dfrac{\pi}{2}\cos x+\sin\dfrac{\pi}{2}\sin x\\&=0\cdot\cos x+1\cdot\sin x=\sin x\end{align*}$
Combining a sine and a cosine
An expression like ${a\sin x+b\cos x}$ is always a single sine curve. The sum formula shows how. Let
${k=\sqrt{a^2+b^2}}$
${a\sin x+b\cos x=k\sin(x+\varphi)}$
where ${\cos\varphi=\dfrac{a}{k}}$ and ${\sin\varphi=\dfrac{b}{k}}$
Here ${\varphi}$ (phi) is an angle. To see why, expand the right side with the sum formula:
$\begin{align*}&k\sin(x+\varphi)\\&=k\cos\varphi\,\sin x+k\sin\varphi\,\cos x\\&=a\sin x+b\cos x\end{align*}$
Example 7: Write ${\sin x+\sqrt{3}\cos x}$ in the form ${k\sin(x+\varphi)}$.
Solution:
Here ${a=1}$ and ${b=\sqrt{3}}$, so
${k=\sqrt{1^2+(\sqrt{3})^2}=\sqrt{4}=2}$
The angle ${\varphi}$ has ${\cos\varphi=\dfrac{1}{2}}$ and ${\sin\varphi=\dfrac{\sqrt{3}}{2}}$. So ${\varphi=\dfrac{\pi}{3}}$, and
$\begin{align*}&\sin x+\sqrt{3}\cos x\\&=2\sin\left(x+\dfrac{\pi}{3}\right)\end{align*}$
This tells us at once that the graph has amplitude ${2}$ and phase shift ${-\dfrac{\pi}{3}}$.
Summary
- Sine formulas mix sine and cosine and keep the sign: ${\sin(A+B)}$ has ${+}$.
- Cosine formulas pair like with like and switch the sign: ${\cos(A+B)}$ has ${-}$.
- Use them to find exact values for angles like ${15^\circ}$ and ${75^\circ}$, and to prove identities.
- ${a\sin x+b\cos x=k\sin(x+\varphi)}$, with ${k=\sqrt{a^2+b^2}}$.