# Cos sin problem?

• Feb 4th 2009, 07:16 PM
view360
Cos sin problem?
Hello. I have a test tomorrow and I'm not sure in one area.

$\displaystyle <\cos \theta, \sin \theta> \cdot <\cos (180+ \theta), \sin(180+ \theta)>$

Explain why the value does not depend on the value of $\displaystyle \theta$.

Could someone please talk me through this step by step? thanks.
• Feb 4th 2009, 07:19 PM
Jhevon
Quote:

Originally Posted by view360
Hello. I have a test tomorrow and I'm not sure in one area.

$\displaystyle <\cos \theta, \sin \theta> \cdot <\cos (180+ \theta), \sin(180+ \theta)>$

Explain why the value does not depend on the value of $\displaystyle \theta$.

Could someone please talk me through this step by step? thanks.

here's something you probably never thought of, take the dot product. what do you get?
• Feb 4th 2009, 07:23 PM
view360
Quote:

Originally Posted by Jhevon
here's something you probably never thought of, take the dot product. what do you get?

$\displaystyle <\cos \theta \cos(180+\theta), \sin \theta \sin(180+ \theta)>$
• Feb 4th 2009, 07:28 PM
Jhevon
Quote:

Originally Posted by view360
cos$\displaystyle theta$cos(180+$\displaystyle theta$), sin$\displaystyle theta$sin(180+$\displaystyle theta$)

that's not the dot product

$\displaystyle \left< a,b \right> \cdot \left< c,d \right> = ac + bd$
• Feb 4th 2009, 07:30 PM
view360
$\displaystyle \cos \theta \cos(180+ \theta) + \sin \theta \sin (180+ \theta)$?

btw is this really calc? I'm only a freshman, isn't calc a much more advanced class?
• Feb 4th 2009, 07:34 PM
Jhevon
Quote:

Originally Posted by view360
cos$\displaystyle theta$cos(180+$\displaystyle theta$) + sin$\displaystyle theta$sin)180+$\displaystyle theta$)?

yes, now recall that $\displaystyle \cos (A - B) = \cos A \cos B + \sin A \sin B$

Quote:

btw is this really calc? I'm only a freshman, isn't calc a much more advanced class?
yeah, you're right. i was tutoring calc 3 today and we were working with dot products, so my mind automatically filed this under calculus :p i'll have it changed.
• Feb 4th 2009, 07:41 PM
view360
Quote:

Originally Posted by Jhevon
yes, now recall that $\displaystyle \cos (A - B) = \cos A \cos B + \sin A \sin B$

yeah, you're right. i was tutoring calc 3 today and we were working with dot products, so my mind automatically filed this under calculus :p i'll have it changed.

Sorry if I sound stupid lol, but i'm not making any connections between the two?
• Feb 4th 2009, 07:44 PM
Jhevon
Quote:

Originally Posted by view360
Sorry if I sound stupid lol, but i'm not making any connections between the two?

replace A with $\displaystyle \theta$ and replace B with $\displaystyle 180 + \theta$

do you see it now?
• Feb 4th 2009, 07:51 PM
view360
Quote:

Originally Posted by Jhevon
replace A with $\displaystyle \theta$ and replace B with $\displaystyle 180 + \theta$

do you see it now?

Oh okay, so it equals cos(A-B), which if you replace it, equals cos(theta-180+theta). So it ends up just equaling cos180, the theta doesn't change that. But why is A = $\displaystyle \theta$ and B $\displaystyle 180 + \theta$?