# Yet another induction proof

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.

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Staff Emeritus
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?

Homework Helper
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?

Homework Helper
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 $$(-1)^{k+1}\cos(x) = (-1) (-1)^k \cos(x)$$

$$LHS=-\cos(k\pi+x)=\cos(k\pi+x+\pi)...$$

ans so on... all too easy

Homework Helper
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,

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Staff Emeritus
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 $\cos(k\pi+x)=(-1)^k\cos(x)$

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