SOLUTION: let f(x) = cos x. Prove that... [f(x+h) - f(x)]/h = cosx[(cosh-1)/h]- sinx[sinh/h] I have no clue how to do this one.

Algebra ->  Trigonometry-basics -> SOLUTION: let f(x) = cos x. Prove that... [f(x+h) - f(x)]/h = cosx[(cosh-1)/h]- sinx[sinh/h] I have no clue how to do this one.      Log On


   

Question 188287: let f(x) = cos x. Prove that...
[f(x+h) - f(x)]/h = cosx[(cosh-1)/h]- sinx[sinh/h]
I have no clue how to do this one.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Note: cos%28x%2Bh%29=cos%28x%29cos%28h%29-sin%28x%29sin%28h%29 this is one of the many trig identities

Since f%28x%29+=+cos%28x%29, this means that f%28x%2Bh%29=cos%28x%2Bh%29=cos%28x%29cos%28h%29-sin%28x%29sin%28h%29 or simply f%28x%2Bh%29=cos%28x%29cos%28h%29-sin%28x%29sin%28h%29


%28f%28x%2Bh%29-f%28x%29%29%2Fh Start with the given expression


%28%28+cos%28x%29cos%28h%29-sin%28x%29sin%28h%29+%29-+cos%28x%29%29%2Fh Plug in f%28x%29+=+cos%28x%29 and f%28x%2Bh%29=cos%28x%29cos%28h%29-sin%28x%29sin%28h%29


%28cos%28x%29cos%28h%29+-+cos%28x%29+-sin%28x%29sin%28h%29+%29%2Fh+ Remove the parenthesis and rearrange the terms (I just moved the last cosine term right after the first term)


%28cos%28x%29+%28cos%28h%29+-+1%29+-sin%28x%29sin%28h%29%29%2Fh Factor the term cos%28x%29 from the first two terms.


%28cos%28x%29+%28cos%28h%29+-+1%29%29%2Fh+-%28sin%28x%29sin%28h%29%29%2Fh Break up the fraction


So where f%28x%29=cos%28x%29


Note: if you are in Calculus, then this would show that the derivative of cos%28x%29 is -sin%28x%29 (with a few more steps).