Question

How would I find the exact value of sin(7pi/8) using a half-angle identity?

Answer 1

7pi/8 is half of 14pi/8 which is 7pi/4. So use the half angle formula and use 7pi/4 for x

Answer 2

How can you change it like that?, because 7pi/8 isn't equal to 7pi/4.. That confuses the crud out of me.

Answer 3

\(\bf {\color{blue}{ \cfrac{7}{8}\cdot 2\implies \cfrac{7}{4}\qquad \cfrac{7\pi}{8}\cdot 2\implies \cfrac{7\pi}{4}}}\\ \quad \\
sin\left(\frac{7\pi}{8}\right)\implies sin\left(\cfrac{\frac{7\pi}{4}}{2}\right)\)

Answer 4

Oh, okay. So the only reason he could do that was because the problem has a two in it?

Answer 5

yes \(\bf sin\left(\frac{7\pi}{8}\right)\implies sin\left(\cfrac{\frac{7\pi}{4}}{2}\right)=\pm\sqrt{\cfrac{1-cos\left(\frac{7\pi}{4}\right)}{2}}\)

Answer 6

Are my answers supposed to be deciamls, or the ones with pi in them? I have no idea how to do this...

Answer 7

well.. what's the \(\bf cos\left(\frac{7\pi}{4}\right)\quad ?\)

Answer 8

.707?

Answer 9

well.. or one can say \(\bf \cfrac{\sqrt{2}}{2}\) check your Unit Circle, thus \(\bf sin\left(\cfrac{\frac{7\pi}{4}}{2}\right)=\pm\sqrt{\cfrac{1-cos\left(\frac{7\pi}{4}\right)}{2}} \implies sin\left(\cfrac{\frac{7\pi}{4}}{2}\right)=\pm\sqrt{\cfrac{1-\frac{\sqrt{2}}{2}}{2}} \\ \quad \\

sin\left(\cfrac{\frac{7\pi}{4}}{2}\right)=\pm\sqrt{\cfrac{\frac{\sqrt{2-\sqrt{2}}}{2}}{2}}\implies sin\left(\cfrac{\frac{7\pi}{4}}{2}\right)=\pm\sqrt{2-\sqrt{2}}\)

Answer 10

I can't see it...

Answer 11

?

Answer 12

refresh the page maybe

Answer 13

It's not that, it goes off the right side too far.

Answer 14

hmm ok... \(\bf sin\left(\cfrac{\frac{7\pi}{4}}{2}\right)=\pm\sqrt{\cfrac{1-cos\left(\frac{7\pi}{4}\right)}{2}} \\ \quad \\\implies sin\left(\cfrac{\frac{7\pi}{4}}{2}\right)=\pm\sqrt{\cfrac{1-\frac{\sqrt{2}}{2}}{2}} \\ \quad \\

sin\left(\cfrac{\frac{7\pi}{4}}{2}\right)=\pm\sqrt{\cfrac{\frac{\sqrt{2-\sqrt{2}}}{2}}{2}}\\ \quad \\\implies sin\left(\cfrac{\frac{7\pi}{4}}{2}\right)=\pm\sqrt{2-\sqrt{2}}\)

Answer 15

Haha. I just did the exact same thing! :p
Okay, two questions. #1) How did you get sqrt2/2? And #2) What is that third step? How did we go from 1-sqrt2/2/2, to sqrt2-sqrt2/2/2???

Answer 16

#1) by using the Unit Circle :), I could have gotten it from the pythagorean theorem anyway

#2) is a fractions addition, so \(\bf 1-\cfrac{\sqrt{2}}{2}\implies \cfrac{1}{1}-\cfrac{\sqrt{2}}{2}\implies \cfrac{2-\sqrt{2}}{2}
\)

Answer 17

I'm a bit lagged

Answer 18

Oh, okay, I get the first part. I still don't get the second. Maybe I need to go back to middle school. How did the top 1 become a 2?

Answer 19

well... that's just a fractions' addition.. you use the LCD for that

Answer 20

Me too. Takes it a million years to catch up on my typing..

Answer 21

http://www.mathsisfun.com/fractions_subtraction.html

Answer 22

Oh. Got it. I promise I am nowhere near as dumb as I seem right now.

Answer 23

How did you use the Unit Circle to get sqrt2/2?

Answer 24

http://www.sciencedigest.org/UnitCircle.gif