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January 16th, 2015, 07:23 PM   #1
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Question 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.
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January 16th, 2015, 07:30 PM   #2
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Quote:
Originally Posted by unistu View Post
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.
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January 16th, 2015, 07:55 PM   #3
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Question

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.
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January 17th, 2015, 02:29 AM   #4
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That's correct.
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January 17th, 2015, 06:07 AM   #5
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Now, do you know what "" and "" are?

Last edited by greg1313; January 17th, 2015 at 11:33 AM.
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January 17th, 2015, 06:36 AM   #6
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Quote:
Originally Posted by Country Boy View Post
"" and ""
$\displaystyle \cos 3\pi$

Last edited by greg1313; January 17th, 2015 at 11:34 AM.
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January 17th, 2015, 01:36 PM   #7
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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.
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January 17th, 2015, 02:00 PM   #8
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$\displaystyle \cos(3\pi + \alpha) = \cos(3\pi)\cos(\alpha) - \sin(3\pi)\sin(\alpha)$

$\displaystyle \cos(3\pi + \alpha) = (-1) \cdot \cos(\alpha) - (0) \cdot \sin(\alpha) = -\cos(\alpha)$
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January 17th, 2015, 03:23 PM   #9
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Quote:
Originally Posted by unistu View Post
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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