This will only be a sketch of the proof, I will leave the demonstration of theOriginally Posted byrabywebb

truth of the base case to the reader, as well as the weasle words needed

to make this a proof by induction.

We are asked to prove:

Let (for convienience only):

and

So we now need to prove that:

Suppose this is true for then we need to prove that it is true

for .

Now:

and:

Thus to prove what is to be proven for is equivalent to proving that:

,

or:

.

Which may be simplified to:

.

To prove this (there must be many other ways) use the trig-identity for

the of a sum to obtain:

.

Now substitute the RHS of for on both the RHS and LHS of ,

on simplification these are both:

.

Which proves:

from the assumption:

.

RonL