Find the exact value cos...sin...

[ScienceJen]

Hello again.

I have been told by my tutor (I'm taking calculus online) is too busy to help me because he has 160 students. I'm so very grateful that I found this site and wish I had found it sooner!

Thank you again.

 

[Dr.Peterson]

Correct: [math]\cos\left(\frac{5\pi}{8}\right)\sin\left(\frac{5\pi}{8}\right)=\frac{1}{2}\sin\left(\frac{5\pi}{4}\right)=\frac{1}{2}\frac{-\sqrt{2}}{2}=\frac{-\sqrt{2}}{4}[/math]
Good work.

 

[ScienceJen]

View attachment 31838
Correct: [math]\cos\left(\frac{5\pi}{8}\right)\sin\left(\frac{5\pi}{8}\right)=\frac{1}{2}\sin\left(\frac{5\pi}{4}\right)=\frac{1}{2}\frac{-\sqrt{2}}{2}=\frac{-\sqrt{2}}{4}[/math]
Good work.

Thank you

 

[ScienceJen]

How about cos^2 (pi/8)? Does that look good, too?
(It is in the attachment, underneath the other question).

 

[Dr.Peterson]

How about cos^2 (pi/8)? Does that look good, too?
(It is in the attachment, underneath the other question).

Please avoid using pdf's; I didn't even notice it had another page (and some of us don't open them).


This is wrong, for at least two reasons: You omitted essential parentheses, which makes what you wrote wrong; and then you did some silly things trying to add fractions.

 

[ScienceJen]

Please avoid using pdf's; I didn't even notice it had another page (and some of us don't open them).

View attachment 31865
This is wrong, for at least two reasons: You omitted essential parentheses, which makes what you wrote wrong; and then you did some silly things trying to add fractions.

Thanks for letting me know - I will not use pdf format in the future.
I understand the reason for the paratheses. As for the fractions: I'm not suppose to find a common denominator and add while finding the exact value. If I were asked to simplify, I guess that's when I would find a common denominator and add the fractions together.
In this case, here is what I figure: 1/2 * (1/sqrt 2 + 1) = 1 over 2 sqrt 2 + 1/2 would be the answer that I get.

 

[Dr.Peterson]

I understand the reason for the paratheses. As for the fractions: I'm not suppose to find a common denominator and add while finding the exact value. If I were asked to simplify, I guess that's when I would find a common denominator and add the fractions together.
In this case, here is what I figure: 1/2 * (1/sqrt 2 + 1) = 1 over 2 sqrt 2 + 1/2 would be the answer that I get.

You could probably leave your answer as [imath]\frac{1}{2}\left(\frac{1}{\sqrt{2}}+1\right)[/imath]; but many teachers expect you to always simplify (which may include rationalizing the denominator). That will be a class rule you need to check.

But when you simplify, you have to do it right! What you did used the common denominator incorrectly. It would be worth your time to do it right, even if you aren't required to simplify, just to make sure you know how to do it. It will be needed! So give it another try, and show your work.

What you're saying here, I think, is that [imath]\frac{1}{2}\left(\frac{1}{\sqrt{2}}+1\right)=\frac{1}{2\sqrt{2}}+\frac{1}{2}[/imath], so you only distributed. That is valid, and can be considered to be all the simplification you need, again depending on class rules.

 

[ScienceJen]

You could probably leave your answer as [imath]\frac{1}{2}\left(\frac{1}{\sqrt{2}}+1\right)[/imath]; but many teachers expect you to always simplify (which may include rationalizing the denominator). That will be a class rule you need to check.

But when you simplify, you have to do it right! What you did used the common denominator incorrectly. It would be worth your time to do it right, even if you aren't required to simplify, just to make sure you know how to do it. It will be needed! So give it another try, and show your work.

What you're saying here, I think, is that [imath]\frac{1}{2}\left(\frac{1}{\sqrt{2}}+1\right)=\frac{1}{2\sqrt{2}}+\frac{1}{2}[/imath], so you only distributed. That is valid, and can be considered to be all the simplification you need, again depending on class rules.

Ok, thank you.