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 January 16th, 2015, 07:23 PM #1 Member   Joined: Dec 2014 From: Queensland Posts: 59 Thanks: 2 Simplify expression Hi, I need to simplify the following. $$\cos\left(3π+α\right)$$ I am unsure of the steps to take to do this. Thanks Last edited by skipjack; January 17th, 2015 at 07:40 PM.
January 16th, 2015, 07:30 PM   #2
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Quote:
 Originally Posted by unistu Hi, I need to simplify the following. $$\cos\left(3π+α\right)$$ I am unsure of the steps to take to do this. Thanks
Use the sum identity for cosine ...

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

Last edited by skipjack; January 17th, 2015 at 07:41 PM.

 January 16th, 2015, 07:55 PM #3 Member   Joined: Dec 2014 From: Queensland Posts: 59 Thanks: 2 Sorry, I'm still unsure what you mean. Do you mean... $$\cos\left(3π+α\right)= \cos\left(3π\right)\cos\left(α\right)-\sin\left(3π\right)\sin\left(α\right)$$ ??? I'm behind as and an online student. Teacher and student interaction is at its minimum this semester. Haven't had a reply in my class forum for ages from the teacher. cou Last edited by skipjack; January 17th, 2015 at 07:43 PM.
 January 17th, 2015, 02:29 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,786 Thanks: 1029 Math Focus: Elementary mathematics and beyond That's correct.
 January 17th, 2015, 06:07 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,097 Thanks: 847 Now, do you know what "$cos(3\pi)$" and "$sin(3\pi)$" are? Last edited by greg1313; January 17th, 2015 at 11:33 AM.
January 17th, 2015, 06:36 AM   #6
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Quote:
 Originally Posted by Country Boy "$cos(3\pi)$" and "$sin(3\pi)$"
$\displaystyle \cos 3\pi$

Last edited by greg1313; January 17th, 2015 at 11:34 AM.

 January 17th, 2015, 01:36 PM #7 Member   Joined: Dec 2014 From: Queensland Posts: 59 Thanks: 2 I've been working with radians as well as degrees, so I assume $$3π$$ would be the radian measure or $$540°$$ but I don't need values. I'm looking on purplemaths website and I see the identity laws mentioned here. If I do $$\cos(3π)-\sin(3π)=-1$$ I still have $$\cos(α)-\sin(α)$$ would the answer be $$-\cos(α)$$ ? I still don't understand how I can simplify this further. Last edited by skipjack; January 17th, 2015 at 07:46 PM.
 January 17th, 2015, 02:00 PM #8 Math Team   Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387 $\displaystyle \cos(3\pi + \alpha) = \cos(3\pi)\cos(\alpha) - \sin(3\pi)\sin(\alpha)$ $\displaystyle \cos(3\pi + \alpha) = (-1) \cdot \cos(\alpha) - (0) \cdot \sin(\alpha) = -\cos(\alpha)$
January 17th, 2015, 03:23 PM   #9
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Quote:
 Originally Posted by unistu I've been working with radians as well as degrees, so I assume $$3π$$ would be the radian measure or $$540°$$ but I don't need values. I'm looking on purplemaths website and I see the identity laws mentioned here. If I do $$\cos(3π)-\sin(3π)=-1$$ I still have $$\cos(α)-\sin(α)$$ would the answer be $$-\cos(α)$$ ? I still don't understand how I can simplify this further.
You are, I hope, aware that $\displaystyle \cos(3 \pi ) = \cos(2 \pi + \pi ) = \cos(\pi)$? (In degrees this is $\displaystyle \cos(540^\circ) = \cos( 360^\circ + 180^\circ ) = \cos(180^\circ)$.)

-Dan

Last edited by skipjack; January 17th, 2015 at 07:47 PM.

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