$\sin θ - \cos θ$

By Anonymous on Tuesday, January 23, 2001 - 01:05 am :

Can somebody please tell me how to graph

\sin θ - \cos θ

I really don't have a clue how to do it. All the help greatly appreciated. Thanks.

By Tim Martin (Tam31) on Tuesday, January 23, 2001 - 01:27 am :

The trick is to rearrange into the form of a single trig expression. It is possible to write anything of the form

A \sin θ + B \cos θ

R \cos (θ + α)

$R \cos (θ + α) = \sin θ - \cos θ$

Expanding using the identity $\cos (A + B) = \cos A \cos B - \sin A \sin B$ :

$R \cos θ \cos α - R \sin θ \sin α = \sin θ - \cos θ$

Therefore $R \sin α = - 1$

$R \cos α = - 1$

$\Rightarrow \tan α = 1$

$\Rightarrow α = π / 4$

$\Rightarrow R \sin (π / 4) = - 1$

$\Rightarrow R = - 2^{1 / 2}$

Putting this together:

$\sin θ - \cos θ = - 2^{1 / 2} \cos (θ + π / 4)$

Hope that makes some sort of sense, ask if you aren't sure about the reasoning.

Tim

By Anonymous on Tuesday, January 23, 2001 - 07:02 pm :

Thanks, it makes sense now!!!