Proving cos(nπ + x) = (-1)^ncos(x) using induction | Induction proof tutorial

[theperthvan]

Homework Statement

Prove the following using induction:
cos(n\pi + x) = (-1)^ncos(x)

The Attempt at a Solution

Let cos(k\pi + x) = (-1)^k cos(x)
True for k=1

Right Hand Side...
(-1)^{k+1}cos(x)
= -(-1)^k cos(x)
= (-1)^k cos(x)

Since (-1)^{k+1}cos(x) = (-1)^k cos(x)...
Not too sure where to go from here.

 

[cristo]

Why not take the LHS cos((k+1)\pi +x)=cos([k\pi +x]+\pi) and use the double angle formula: cos(a+b)=cosa cosb-sina sinb?

 

[matt grime]

Are you claiming that's a proof for k=1, an attempt at a proof for k=1, or are you stating it is true for k=1 without any proof?

Anyway, this is about proof by induction, and obviously the left hand side is the thing you're going to want to attack inductively (since there is an obvious way to write k+1 as k plus 1 and you know you're trig formulae (cos(A+B)=cos(A)cos(B)-sin(A)sin(B) for instance...)

 

[theperthvan]

I did the LHS, and got

cos((k+1)\pi +x) = cos(k\pi +x)

and from before,

(-1)^{k+1}cos(x) = (-1)^k cos(x)

And so from that, and the initial statement, does that prove it?

 

[mjsd]

I did the LHS, and got

cos((k+1)\pi +x) = cos(k\pi +x)

and from before,

(-1)^{k+1}cos(x) = (-1)^k cos(x)

And so from that, and the initial statement, does that prove it?

err.. isn't (-1)^{k+1}\cos(x) = (-1) (-1)^k \cos(x)

so using your assumption:
LHS=-\cos(k\pi+x)=\cos(k\pi+x+\pi)...

ans so on... all too easy

 

[matt grime]

I did the LHS, and got

cos((k+1)\pi +x) = cos(k\pi +x)

this is wrong: look at the graph of cos(x).

and from before,

(-1)^{k+1}cos(x) = (-1)^k cos(x)

And so from that, and the initial statement, does that prove it?

this is also wrong: you've just said 1=-1.

 

[theperthvan]

this is wrong: look at the graph of cos(x).
this is also wrong: you've just said 1=-1.

Oops, silly mistake.

So let me try this again,

cos(k\pi + x) = (-1)^kcos(x)

RHS
(-1)^{k+1}cosx = -(-1)^{k}cosx

LHS
cos[(k+1)\pi + x] = cos[(k\pi + x) + \pi]
using double angle formula,
= -cos(k\pi + x)

And so, cos[(k+1)\pi + x] = (-1)^{k+1}cosx
=> -cos(k\pi + x) = -(-1)^{k}cosx

kill the negatives

cos(k\pi + x) = (-1)^{k}cosx

Cheers,

 

[cristo]

Oops, silly mistake.

So let me try this again,

cos(k\pi + x) = (-1)^kcos(x)

RHS
(-1)^{k+1}cosx = -(-1)^{k}cosx

LHS
cos[(k+1)\pi + x] = cos[(k\pi + x) + \pi]
using double angle formula,
= -cos(k\pi + x)

And so, cos[(k+1)\pi + x] = (-1)^{k+1}cosx
=> -cos(k\pi + x) = -(-1)^{k}cosx

Here, you've assumed that the statement holds for n=k+1, and used this to prove for n=k. This to me doesn't seem right, as surely you need to show that the statement holding for n=k => the statement is true for n=k+1.

To do this, we suppose that \cos(k\pi+x)=(-1)^k\cos(x) is true. Now, we use this to prove that it hold for n=k+1; \cos([k+1]\pi+x)=\cos([k\pi+x]+\pi)=-\cos(k\pi+x)=-(-1)^k\cos(x)=(-1)^{k+1}\cos(x) , where we note that the penultimate equality holds from our inductive hypothesis that \cos(k\pi+x)=(-1)^k\cos(x)

 

[matt grime]

What are you doing? LHS and RHS of what? Not what you wrote one line above.

Induction proofs are taught by rote. So reproduce the style of proof.

And don't write things going from both sides like this. Sure, that's often how you work out what's going on (to get from A to B yo go from A to C and from B to C as well). What you write should be a simple argument with no jumping around:

It's trivially true for n=1.

Consider cos(pi(k+1)+x), and simplify it and now use the inductive hypothesis that the result is true for k (and everything between 1 and k as well if necessary - it isn't in this case).