# Finding Exact Value

• Dec 8th 2007, 05:58 AM
AnTh
Finding Exact Value
Good morning all!

There is a problem that I am unable to solve:

Find the exact value of cos[3(pi)/8].

I can't figure out how to solve that for exact solution. My teacher said to use a double-angle identity but I am very confused. My class is new to all this. So any help regarding this problem would be appreciated. Also if it isn't asking to much, can you show the steps if you are able to help me.

Thanks, Anth :)
• Dec 8th 2007, 06:32 AM
Krizalid
Quote:

Originally Posted by AnTh
Find the exact value of cos[3(pi)/8].

Well, this method is not so nice at all, but it's something.

Start by showing that $\cos \frac{\varphi }
{2} = \frac{{\sqrt {1 + \cos \varphi } }}
{\sqrt2}.$

So

$\cos \frac{3}
{8}\pi = \cos \frac{1}
{2} \cdot \frac{3}
{4}\pi = \frac{{\sqrt {1 + \cos \dfrac{3}
{4}\pi } }}
{{\sqrt 2 }}.$

Now the remaining challenge is to compute $\cos\frac34\pi.$

Note that

$\cos \frac{3}
{4}\pi = \cos \left( {\pi - \frac{\pi }
{4}} \right) = \cos \pi \cos \frac{\pi }
{4} + \sin \pi \sin \frac{\pi }
{4} = - \frac{{\sqrt 2 }}
{2}.$

And finally

$\cos \frac{3}
{8}\pi = \frac{{\sqrt {1 - \dfrac{{\sqrt 2 }}
{2}} }}
{{\sqrt 2 }} = \frac{{\sqrt {2 - \sqrt 2 } }}
{2}\,\blacksquare$
• Dec 8th 2007, 06:59 AM
AnTh
Quote:

Originally Posted by Krizalid
Well, this method is not so nice at all, but it's something.

Start by showing that $\cos \frac{\varphi }
{2} = \frac{{\sqrt {1 + \cos \varphi } }}
{\sqrt2}.$

Well thanks alot for answering this quickly. Yes, your method works but im a little confused with the first formula you got and use (shown above ^)as I've never learned that. However, it works which is great. Thanks for helping out. But if there are any other methods which is more simple that would be great. My teacher mentioned using the double angle identities so if there is any method using that, that would be great if anyone can help out. But i would like to thank you Krizalid. If worse comes to worse, I'll use your method and my teacher would be impressed.

Thanks, Anth:)
• Dec 8th 2007, 07:12 AM
Krizalid
Well, my method is not hard to understand, but the first formula can be derived as follows:

Consider the double-angle cosine formula $\cos2u=\cos^2u-\sin^2u.$

So $\cos2u=\cos^2u-(1-\cos^2u)=2\cos^2u-1.$

Substitute $u=\frac\varphi2,$ after that make $\cos\frac\varphi2$ the subject of the formula and you're done.

--

I dunno if the method by applyin' the double-angle cosine formula leads an easy or hard way to compute what you want. (Since I haven't tried.)
• Dec 8th 2007, 07:25 AM
AnTh
Thank you. You made it a bit more clear to me. You are right when you say that your method isn't to hard to understand. I get most of it except the first part. I dont understand the concept of http://www.mathhelpforum.com/math-he...5c2ba2e8-1.gif Honestly, I've never seen that before as I don't think my class has learned it so i don't understand the concept of that:o. That is why I'm asking if there are other methods, but I appreciate your response. :)
• Dec 8th 2007, 07:33 AM
Krizalid
Haha, no problem!

The only reason to use such formula is that

$\frac{3}
{8}\pi = \frac{1}
{2} \cdot \frac{3}
{4}\pi = \frac{{\dfrac{3}
{4}\pi }}
{2}.$

Then plug that into the aforesaid formula. Now, why did I do this? 'cause I knew that evaluating $\cos \frac{3}
{4}\pi$
leads an easy computation. (See post #2.)

Since $\cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta ,$ and since we're workin' with a nice angle $\frac\pi4,$ the expected value will be easy to handle.

The rest are just arithmetic calculations.
• Dec 8th 2007, 07:48 AM
AnTh
Thank you. I pretty much get it. Im still a bit confused on the whole concept but ill try to make sense of it when i write it out in good. I appreciate your help. But ill still accept any other methods if anyone can think of any.

Thanks, Anth:)
• Dec 8th 2007, 08:06 AM
Plato
Quote:

Originally Posted by AnTh
But if there are any other methods which is more simple that would be great. My teacher mentioned using the double angle identities so if there is any method using that

$
\begin{array}{l}
\cos (2A) = 2\cos ^2 (A) - 1 \\
\cos \left( {\frac{{3\pi }}{4}} \right) = 2\cos ^2 \left( {\frac{{3\pi }}{8}} \right) - 1 \\
- \frac{{\sqrt 2 }}{2} = 2\cos ^2 \left( {\frac{{3\pi }}{8}} \right) - 1 \\
\cos \left( {\frac{{3\pi }}{8}} \right) = \sqrt {\frac{1}{2} - \frac{{\sqrt 2 }}{4}} \\
\end{array}$
• Dec 8th 2007, 08:15 AM
AnTh
Thank you so much. I understand it all now and that answer is exactly what im looking for. Thank you to the both of you. I appreciate it alot. I'm new to this community and seem to enjoy it already. Very welcoming and helpful.:)