Trig Identities?

Question

I'm having trouble solving my last question on my summer assignment:

Derive sin2x + cos2x = 1

with only using

cos(x + y) = (cosx)(cosy) - (sinx)(siny)
sin(-x) = -sinx
cos(-x) = cosx
cos0 = 1

Please explain how to solve the problem. Thank You.

Answer 1

I assume you meant:
Derive sin^2 x + cos^2 x = 1

**********
cos^2 x + sin^2 x
= (cos x)^2 + (sin x)^2
= (cos x)(cos x) + (sin x)(sin x)
= (cos x)(cos x) - (sin x)(- sin x)
= (cos x)(cos -x) - (sin x)(sin -x)
= cos (x + (-x))
= cos 0
= 1

Answer 2

I find the best thing to do when you have to memorize a lot of formulas is to make some flashcards. Within just a few sessions, one can easily memorize 15-20 formulas. I once used that method to memorize over 130 formulas for an actuary exam. Another trick: take the derivative of some of your functions. You should be able to calculate the derivative of tanx by translating to sinx and cosx and applying the quotient rule. One more trick: use your graphing calc. to your advantage. If you have a TI-89 you shouldn't have any problems as that calculator can compute most integrals explicitly. But, if you have a TI-83 or TI-84 you can still use numerical integral to check your answers. I find it amazing how many students don't use their graphing calculators to their full advantage. I've had many perfect tests, because I checked my work so carefully with a graphing calc. Final trick: when you're given a bunch of tanx's and secx's you can always convert these to sinx and cosx, and you should know what the integral of sinx and cosx are, and be able to use a reduction formula if needed.

Answer 3

First of all, sin(2x) + cos(2x) =1 is not an identity.
I think you meant (sinx)^2 + (cosx)^2 =1.
If you take the relationship a^2 + b^2 = c^2 and divide by c^2 you get:
(a/c)^2 + (b/c)^2 = (c/c)^2
which is:
sin^2 + cos^2 =1

Answer 4

That should be sin^2(x) + cos^2(x) = 1