A Quest for Calculus: October 2009

I intend for this to be the last spit of trig proofs for a while. After this post, we'll have the tools we need to start working on derivatives. Specifically, we're going to work out how to expand $sin{(c+h)}$ and $cos{(c+h)}$ .

Once again, we're employing our now familiar unit circle:

You'll notice I have a fair amount of mark up on there. It seems messy, and there may vary well be an easier way to do it, but this was the way I did it, and it makes sense to me. So, we'll start like always with what we know about the problem:

$\overline{AB}=1$

$\sin{h}=\overline{BE}$

$\cos{h}=\overline{AE}$

$\sin{(c+h)}=\overline{BD}$

$\cos{(c+h)}=\overline{AD}$

There are also some things we can infer about the diagram:

$\angle{ADG}$ is a right angle meaning $\angle{AGD}=\frac{\pi}{2}-\angle{c}$

$\angle{AGB}=\pi-\angle{AGD}=\pi-(\frac{\pi}{2}-\angle{c})=\frac{\pi}{2}+\angle{c}$

$\angle{BGE}=\pi-\angle{AGB}=\pi-(\frac{\pi}{2}+\angle{c})=\frac{\pi}{2}-\angle{c}$

$\angle{GBE}=\frac{\pi}{2}-\angle{BGE}=\frac{\pi}{2}-(\frac{\pi}{2}-\angle{c})=\angle{c}$

Now we're ready to begin expanding our identities. Starting with $\sin{(c+h)}$ :

$\sin{(c+h)}=\overline{BD}=\overline{BC}+\overline{CD}$

We'll start by finding the value for $\overline{BC}$ :

$\cos{(\angle{GBE})}=\cos{(\angle{CBE})}=\cos{c}=\frac{{BC}}{{BE}}=\frac{{BC}}{\sin{h}}$

Simplifying, we get: $\overline{BC}=\cos{c}\sin{h}$

We're halfway there. Now we have to find $\overline{CD}$ :

$\overline{CD}=\overline{EF}$

$\sin{c}=\frac{{EF}}{{AE}}=\frac{{EF}}{\cos{h}}=\frac{{CD}}{\cos{h}}$

Simplifying again, we get: $\overline{CD}=\sin{c}\cos{h}$

So: $\sin{(c+h)}=\overline{BC}+\overline{CD}=\cos{c}\sin{h}+\sin{c}\cos{h}$

So, there's that. Time to move on to $\cos{(c+h)}$ . Much like with $\sin{(c+h)$ , we'll be using a divide and conquer approach.

$\cos{(c+h)}=\overline{AD}=\overline{AF}-\overline{DF}$

So, starting with $\overline{AF}$ :

Thinking back to our days of SOHCAHTOA we know that $\tan{c}=\frac{EF}{AF}$

We also know that $\tan{c}=\frac{\sin{c}}{cos{c}}$ and (from our previous proof) $\overline{EF}=\sin{c}\cos{h}$

So, $\frac{\sin{c}}{\cos{c}}=\frac{\sin{c}\cos{h}}{AF}$

Simplifying we get: $\overline{AF} = \cos{c}\cos{h}$

Again, we find ourselves halfway there. Now we need to find $\overline{DF}$ :

We know $\overline{DF}=\overline{CE}$

We know from our previous example that $\angle{BCE}=\angle{c}$

This tells us $\tan{c}=\frac{CE}{BC}$

Expanding using our previous proofs we get $\frac{\sin{c}}{\cos{c}}=\frac{CE}{\cos{c}\sin{h}}$

Simplifying we get $\overline{DF}=\overline{CE}=\sin{c}\sin{h}$

Now we put it all together: $\cos{(c+h)}=\overline{AF}-\overline{DF}=\cos{c}\cos{h}-\sin{c}\sin{h}$

So, we can now use the following two identities:

$\sin{(c+h)}={\cos{c}\sin{h}+\sin{c}\cos{h}$

$\cos{(c+h)}={\cos{c}\cos{h}+\sin{c}\sin{h}$

Thus, we close the book on trig for a while.

It's time to look at one of the most fundamental identities in trig: $\sin^2x+\cos^2x=1$
I can't count the number of times I've used that identity. I can however, count the number of times I've used it since I knew why it was true. This proof will be a breeze after the SOHCAHTOA work. We'll use the unit circle again for this proof: