Trigonometric Matrixes?

**El Joey** · December 16th 2008, 07:03 AM

I need to prove by induction that this 2x2 matrix:

[cosθ -senθ; senθ cosθ]^n is equal to this 2x2 matrix:

[cos(n)θ -sen(n)θ; sen(n)θ cos(n)θ] for all n ∈ Z.

Since it's by induction, I've already got the $n = 1$ and $n = k$ , but... now what?

Thanks!

**Isomorphism** · December 16th 2008, 07:13 AM

Originally Posted by El Joey

I need to prove by induction that this 2x2 matrix:

[cosθ -senθ; senθ cosθ]^n is equal to this 2x2 matrix:

[cos(n)θ -sen(n)θ; sen(n)θ cos(n)θ] for all n ∈ Z.

Since it's by induction, I've already got the $n = 1$ and $n = k$ , but... now what?

Thanks!

Here's the idea:

Let $Q(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin \theta & \cos\theta \end{pmatrix}$

Show that $Q(\theta)Q(\alpha) = Q(\theta + \alpha)$ , by multiplying out.

Now choose $\alpha = n\theta$ and use the induction hypothesis to finish the proof

**Moo** · December 16th 2008, 07:14 AM

Hello,

Originally Posted by El Joey

I need to prove by induction that this 2x2 matrix:

[cosθ -senθ; senθ cosθ]^n is equal to this 2x2 matrix:

[cos(n)θ -sen(n)θ; sen(n)θ cos(n)θ] for all n ∈ Z.

Since it's by induction, I've already got the $n = 1$ and $n = k$ , but... now what?

Thanks!

Let $A=\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$

Assume that $A^k=\begin{pmatrix} \cos k \theta & - \sin k \theta \\ \sin k \theta & \cos k \theta \end{pmatrix}$

Then $A^{k+1}=A^kA=\begin{pmatrix} \cos k \theta & - \sin k \theta \\ \sin k \theta & \cos k \theta \end{pmatrix} \begin{pmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$

Now make the multiplication and use these formula :
$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$
$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$

and you're done

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