## Sine FormulasEdit

The addition formula for sines is as follows:

$\displaystyle\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

This is an important tool that allows us to relate the sines and cosines of angles of different sizes.

There is a related formula for cosines, discussed in the next section:

$\displaystyle\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

 Worked Example: Sine of $\displaystyle 75^\circ$ You are told that: $\displaystyle\sin(45^\circ)\approx 0.7071$ $\displaystyle\cos(45^\circ)\approx 0.7071$ $\displaystyle\sin(30^\circ)= 0.5$ $\displaystyle\cos(30^\circ)\approx 0.866$ Use the sine addition formula to calculate $\displaystyle\sin(75^\circ)$ Answer: Using the first formula: $\displaystyle\sin(45^\circ+30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$ $\displaystyle\sin(45^\circ+30^\circ) \approx 0.7071\times 0.866 + 0.7071\times 0.5$ $\displaystyle\sin(75^\circ) \approx 0.6123 + 0.3536 \approx 0.966$

 Exercise: Check the worked example Check the answer in the worked example agrees with the correct value by using a calculator.
 Exercise: Sines and Cosines of $\displaystyle 20^\circ, 30^\circ, 40^\circ$ You are told that $\displaystyle\sin(10^\circ)\approx 0.1736$ and $\displaystyle\cos(10^\circ)\approx 0.9848$ Use the formulas to calculate the sine and cosine of $\displaystyle 20^\circ, 30^\circ, 40^\circ$ Check that your answers agree with the values for sine and cosine given by using your calculator to calculate them directly.

The addition formulas are very useful.

Here is a geometric proof of the sine addition formula. The proof also shows how someone could have discovered it.

## ProofEdit

We want to prove:

$\displaystyle\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

First, a word about the diagram used in the proof. How on earth would you come up with a diagram like that?

Well,

• We need a diagram with right triangles and we need to show an angle of $\displaystyle(\alpha+\beta)$, so having $\displaystyle \triangle ABC$ is a must.
• We want to express the lengths in this triangle in terms of lengths of two right triangles, one with angle $\displaystyle\alpha$ and one with angle $\displaystyle\beta$, so adding points like $\displaystyle E$ and $\displaystyle F$ is essential.
• Having got that far we could start trying to solve the problem, and we'd find we ran into a problem when calculating the distance $\displaystyle \overline{BC}$. That's why we split $\displaystyle \overline{BC}$ into $\displaystyle \overline{BD}$ and $\displaystyle \overline{DC}$. We can calculate the distance $\displaystyle \overline{DC}$. It is the same length as $\displaystyle \overline{EF}$. Also $\displaystyle \overline{BD}$ is a length we can calculate using Soh-Cah-Toa.

Be aware that there is nothing really special about the diagram we chose. It's possible, for example, to calculate $\displaystyle\sin(\alpha + \beta)$ using a diagram where the right triangle $\displaystyle \triangle ABE$ has its right angle at $\displaystyle B$ rather than at $\displaystyle E$. You might like to try that.

We've chosen this digram and this lettering because it is exactly the same as used in the Khan Academy video on proving addition formula for sine so if you have trouble with the proof presented here, you can follow it on video instead.

There is a video of the proof which may be easier to follow at the Khan Academy.

### The ProofEdit

First check that the $\displaystyle \angle DBE$ really is the same as $\displaystyle \angle FAE$. That's going to be important to the proof. We're just using the fact that angles in a triangle add up to 180 to make that check, noting that we know the 90 degree angles.

Now an expression for $\displaystyle\sin(\alpha + \beta)$. Here we're using Soh-Cah-Toa. We're going to be using Soh-Cah-Toa a lot.

$\displaystyle\sin(\alpha + \beta) = \frac{\overline{BC}}{\overline{AB}}$

Looking at the diagram we can replace $\displaystyle\overline{BC}$ by $\displaystyle\overline{BD} + \overline{DC}$ and we also have $\displaystyle\overline{DC}=\overline{EF}$ so:

$\displaystyle\sin(\alpha + \beta) = \frac{\overline{BD}+\overline{DC}}{\overline{AB}}=\frac{\overline{BD}+\overline{EF}}{\overline{AB}}= \frac{\overline{BD}}{\overline{AB}} + \frac{\overline{EF}}{\overline{AB}}$

Let's work out another way to express $\displaystyle \overline{BD}$ and another way to express $\displaystyle \overline{EF}$. You'll need to look at the diagram to see which triangles we are using.

 An expression for $\displaystyle \overline{BD}$ $\displaystyle \overline{BD}=\overline{BE}\cos\alpha$ and $\displaystyle \overline{BE}=\overline{AB}\sin\beta$ so $\displaystyle \overline{BD}=\overline{AB}\sin\beta\cos\alpha$ An expression for $\displaystyle \overline{EF}$ $\displaystyle \overline{EF}=\overline{AE}\sin\alpha$ and $\displaystyle \overline{AE}=\overline{AB}\cos\beta$ so $\displaystyle \overline{EF}=\overline{AB}\cos\beta\sin\alpha$

### Putting it all TogetherEdit

$\displaystyle\sin(\alpha + \beta) = \frac{\overline{BD}}{\overline{AB}} + \frac{\overline{EF}}{\overline{AB}} = \frac{\overline{AB}\sin\beta\cos\alpha}{\overline{AB}} + \frac{\overline{AB}\cos\beta\sin\alpha}{\overline{AB}}$

The $\overline{AB}$'s cancel.

$\displaystyle\sin(\alpha + \beta) = \sin\beta\cos\alpha + \cos\beta\sin\alpha = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

We're done!

## ExercisesEdit

 Exercise: Make one of the sides 'one' When we drew the diagram we said nothing about its size. That means we could still choose to make one of the sides be of whatever length we like. We can do this for just one edge. Once we've done that all the other sides lengths are determined. Fixing one length to be a nice value can shorten the proof. So, let us decide that $\displaystyle AB$ is 1 km. Actually we'll not worry about the units whether km, m or cm and just write '1'. $\displaystyle \triangle ABE$ is a right triangle and: $\displaystyle \overline{AE} = \overline{AB} \cos\beta = \cos\beta$ Your task is to simplify the entire proof of the addition formula by replacing the lengths like $\displaystyle \overline{AE}$ with the actual values assuming that we've set $\displaystyle AB=1$. You're effectively removing $\displaystyle AB$ and multiplication and division by $\displaystyle AB$ from the proof. It should become a lot shorter and clearer. You should also mark up the lengths on the diagram, assuming $\displaystyle AB=1$.
 Exercise: Use a different diagram Read the description About the Diagram of how the diagram was constructed again. Make your own diagram that is different to the one shown, with right angles in different places to the diagram shown, and do the proof using it instead. Tip: If you find yourself adding lots of lines and lots of extra points then you are probably making the proof a lot harder than you need to. You want to add only enough lines to be able to 'chase' lengths from one place to another. Once you've got the three basic triangles, one extra line should be enough.