# Need help on a cos/sin explanation problem!

• Jan 20th 2009, 07:12 PM
pyrosilver
Need help on a cos/sin explanation problem!
Explain why the value of [cos$\displaystyle theta$, sin$\displaystyle theta$) *dot* [(cos(90+$\displaystyle theta$), sin(90+$\displaystyle theta$)] is independent of $\displaystyle theta$.

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^^ this is one of the problems I have due tomorrow. Problem is I don't even know what it's asking. independent? Are they saying that if you dot those, you can remove the theta and it still works? If this is what its asking I have no idea. i've been thinking about this for like half an hour and i think I'm missing something key? Please help =(
• Jan 20th 2009, 07:18 PM
Prove It
Quote:

Originally Posted by pyrosilver
Explain why the value of [cos$\displaystyle theta$, sin$\displaystyle theta$) *dot* [(cos(90+$\displaystyle theta$), sin(90+$\displaystyle theta$)] is independent of $\displaystyle theta$.

--

^^ this is one of the problems I have due tomorrow. Problem is I don't even know what it's asking. independent? Are they saying that if you dot those, you can remove the theta and it still works? If this is what its asking I have no idea. i've been thinking about this for like half an hour and i think I'm missing something key? Please help =(

Take the dot product and remember that

$\displaystyle \cos{(x - y)} = \cos{x}\cos{y} + \sin{x}\sin{y}$.

You should find that you get something that doesn't involve a $\displaystyle \theta$.
• Jan 20th 2009, 07:19 PM
Mush
Quote:

Originally Posted by pyrosilver
Explain why the value of [cos$\displaystyle theta$, sin$\displaystyle theta$) *dot* [(cos(90+$\displaystyle theta$), sin(90+$\displaystyle theta$)] is independent of $\displaystyle theta$.

--

^^ this is one of the problems I have due tomorrow. Problem is I don't even know what it's asking. independent? Are they saying that if you dot those, you can remove the theta and it still works? If this is what its asking I have no idea. i've been thinking about this for like half an hour and i think I'm missing something key? Please help =(

If this is supposed to be the dot product of two vectors $\displaystyle (\cos(\theta), \sin(\theta))$ and $\displaystyle (\cos(\theta + 90), \sin(\theta + 90))$, then:

$\displaystyle (\cos(\theta), \sin(\theta)) \cdot (\cos(\theta + 90), \sin(\theta + 90)) $$\displaystyle = \cos(\theta)\times\cos(\theta + 90)+sin(\theta)\times \sin(\theta + 90) \displaystyle = \cos(\theta)(\cos(\theta)\cos(90)-\sin(\theta)\sin(90))+sin(\theta)( \sin(\theta)\cos(90) + \sin(90)\cos(\theta)) Since \displaystyle \sin(90) = 1 and \displaystyle \cos(90) = 0 : \displaystyle = \cos(\theta)(-\sin(\theta))+sin(\theta)\cos(\theta) \displaystyle = -\cos(\theta)\sin(\theta)+sin(\theta)\cos(\theta) = 0 You should expect this result, since the 2nd vector is the same as the first vector, except it has been rotated through and angle of 90 degrees, and is hence perpendicular. And the dot product of two perpendicular vectors is always 0. • Jan 20th 2009, 07:25 PM Prove It Quote: Originally Posted by Mush If this is supposed to be the dot product of two vectors \displaystyle (\cos(\theta), \sin(\theta)) and \displaystyle (\cos(\theta + 90), \sin(\theta + 90)) , then: \displaystyle (\cos(\theta), \sin(\theta)) \cdot (\cos(\theta + 90), \sin(\theta + 90))$$\displaystyle = \cos(\theta)\times\cos(\theta + 90)+sin(\theta)\times \sin(\theta + 90)$

$\displaystyle = \cos(\theta)(\cos(\theta)\cos(90)-\sin(\theta)\sin(90))+sin(\theta)( \sin(\theta)\cos(90) + \sin(90)\cos(\theta))$

Since $\displaystyle \sin(90) = 1$ and $\displaystyle \cos(90) = 0$:

$\displaystyle = \cos(\theta)(-\sin(\theta))+sin(\theta)\cos(\theta)$

$\displaystyle = -\cos(\theta)\sin(\theta)+sin(\theta)\cos(\theta) = 0$

You should expect this result, since the 2nd vector is the same as the first vector, except it has been rotated through and angle of 90 degrees, and is hence perpendicular. And the dot product of two perpendicular vectors is always 0.

My way is easier :P
• Jan 20th 2009, 07:29 PM
pyrosilver
Thanks both of you =) I understand it now!
• Jan 20th 2009, 08:18 PM
Mush
Quote:

Originally Posted by Prove It
My way is easier :P

Yeah. Way easier!