need help with sin & cos identities?

Question

I’m looking over Derivatives of Sines and Cosines and can’t figure out how to go from step #1 to step #2. The author of my calculus book has a habit of not tell you how he goes from one step to the other. I have the sin/cos identities in front of me and I still can’t figure it out.

Step #1
Lim h-->0 [ (sinxcosh + sinhcosx) – sinx ] / h

Step #2
Lim h-->0 [ (cosx)(sinh/h) – (sinx)(1–cosh/h) ]

the x's should be h's
sorry

ignore that last detail, i'm an idiot

Answer 1

They are not identities. The author factored.

You have: [(sinxcosh + sinhcosx) - sin x] /h

common denominator gives:
(sinxcosh)/h +(sinhcosx)/h - sinx/h

so factor out a sin x to get
sinx (cosh-1)/h + (sinhcosx)/h

Factor out a negative in first term gives:
-sinx (-cosh+1)/h +(sinhcosx)/h

That is what Step 2 is. it is basically distribution and rearranging. There are no trig identities used.

Answer 2

Derivative of sin x
------------------------
f `(x) = lim h->0 (sin(x + h) - sin x) / h
= lim h->0 [ sin x cos h + cos x sin h - sin x ] / h
= sinx - sinx + lim h->0 (sin h / h) (cos x)
= cos x

Derivative of cos x is derived in a similar manner.