I thought I’d post this quick simplification using trig identities since it gave me a bit of a run for my money. Partially so I remember it better and second, to lend a bit of help to those who might be stuck.
The Problem: Simplify
Now, how hard could this be?!? As it turns out, it’s not too bad, but the trick is in factoring what we have.
Factor things out:
Great! Now that we have this in better form we need to bring in a couple of trig identities which will help us solve this and one that I’m going to cite because it could trick you!
Pythagorean Identity 1:
Pythagorean Identity 2:
Pythagorean Identity 3: (This is the tricky one! As much as it seems like we could use this identity we would end up loosing terms. Don’t use this)
Now, things should start looking a bit simpler to solve. So, here we go starting with what we factored:
1)
2) Use Identity 1 to substitute for the first term:
3) Use Identity 2 to substitute for the first part of the second term:
4) Now combine the two like terms:
5) And so we have it! The final result is:
Now, looking back, if you notice in identity 3, if you use it, you end up with a term which multiplied out you loose the negative and the 2 in front of cos. This is a big problem because if you did it this way, what you get in the end does not equal what you had in the beginning and thus, isn’t a valid way to simplify this problem. As a result, you must use identities 1 and 2 as shown in the 5 steps above.
Cheers!