Is there a fast way in Verifying Trigonometric Identities?

Question

Hi,

I taking pre-calculus right now and its driving me crazy!!! The section were on (verifying trigonometric identities) is making my head hurt. Is there a simple way to do them? My pre-calc teacher is saying that i should start with the complicated side and work it through. The problem is that i don't understand how to get what. My homework is driving me crazy HELP!.

If anyone is out there that can explain how to do this, I really appreciate your help. Thank you.

Answer 1

Heres the way I do them, which isn't always the simplest, but always works.
Turn the most complicated side into something which *only* involves sin or cos.
Turn any "non-simple" sin's or cos's into simpler ones - eg using the double-angle formula, or turning sums into products or vice versa.
Simplify everything, putting all fractions over a common denominator.
Use the fact that (sin x)^2 + (cos x)^2 = 1 to simplify things even more.

Now, if you still can't see how to make this equal to the other side, do exactly the same process on the other (simpler) side.

If they *still* aren't equal, look at the places they differ, and use (sin x)^2 + (cos x)^2 = 1 again to change any sin's or cos's to cos's or sin's to make everything match up.

If you see sums on one side and products on the other, you may need to use product/sum trig identities.

I haven't yet found a trig identity that isn't provable using that technique.

Answer 2

I would write them all on a notecard so you can look at them while you do your homework. (I had a notecard with ALL my "need to know" stuff on it - it helped so much!)

The thing that I found to work really well was to make everything into sines and cosines. Usually this allowed me to cancel things out. (My calculus teacher always said that when it doubt, go to sine and cosine - they are the easiest to work with!)

Answer 3

i have some mcgraw hill books such as "precalculus demystified", and "calculus demystified"
I think you can download them with eMule they have very clear explanations
The second book explains the trigonometric identities

Answer 4

Yeah I feel you.

I'm in AP Calculus.

You will get through it, here's what I did:

Notecard, in the back of a translucent-backed- graphing calculator =)

Answer 5

(cos x + sin x) (a million - cos x sin x) distribution: cos(x) - cos^2(x)sin(x) + sin(x) - cos(x)sin^2(x) changing cos^2(x) and sin^2(x) using the identity sin^2(x) + cos^2(x) = a million (i.e. sin^2(x) = a million-cos^2(x)) cos(x) - [a million-sin^2(x)]*sin(x) + sin(x) - cos(x)*[a million-cos^2(x)] multiplying out: cos(x) - [sin(x)-sin^3(x)] + sin(x) - [cos(x)-cos^3(x)] Simplifying (removing brackets): cos(x) - sin(x) + sin^3(x)] + sin(x) - cos(x) + cos^3(x) Rearranging: cos(x) - cos(x) +sin(x) - sin(x) + sin^3(x) + cos^3(x) = sin^3(x) + cos^3(x) So we've shown that by basically alternative using familiar identity that: sin^3x + cos^3x = (cos x + sin x) (a million - cos x sin x)