Author Topic: pi/8 (Read 1423 times) Tweet Share

sachinmachin · « **on:** September 30, 2009, 07:11:03 pm »

$\frac{\pi }{8}$

what 2 angles produce this for a compound angle formula??

kurrymuncher · « **Reply #1 on:** September 30, 2009, 07:13:35 pm »

If your wondering how to break this up, dont, because you cant. Im guessing this should be used in a tan or sin double angle formula, so make Pi/8 equal Thetha, and then use the formulas.

dejan91 · « **Reply #2 on:** September 30, 2009, 07:15:42 pm »

Double angle forumula...?

EDIT: ...didn't read kurrymuncher's post.

TrueTears · « **Reply #3 on:** September 30, 2009, 07:15:55 pm »

If you're trying to evualate $\cos(\frac{\pi}{8})$ or $\sin(\frac{\pi}{8})$ , I'd use the half angle formula.

Ie, $\sin(\frac{\pi}{8}) = \sin(\frac{1}{2} \times \frac{\pi}{4})$

sachinmachin · « **Reply #4 on:** September 30, 2009, 07:20:05 pm »

Quote from: kurrymuncher on September 30, 2009, 07:13:35 pm

If your wondering how to break this up, dont, because you cant. Im guessing this should be used in a tan or sin double angle formula, so make Pi/8 equal Thetha, and then use the formulas.

twas in heffernan 09 exam 1 which we did as a trial exam at school. i was convinced that it was a compound angle formula question as it asked:

Show that $cos(\frac{\pi }{8}) =$ something, lol i forgot. (it was a surd of some sort)

sachinmachin · « **Reply #5 on:** September 30, 2009, 07:21:46 pm »

Quote from: TrueTears on September 30, 2009, 07:15:55 pm

If you're trying to evualate $\cos(\frac{\pi}{8})$ or $\sin(\frac{\pi}{8})$ , I'd use the half angle formula.

Ie, $\sin(\frac{\pi}{8}) = \sin(\frac{1}{2} \times \frac{\pi}{4})$

yeh i suppose your right TT. i really do suck:(

TrueTears · « **Reply #6 on:** September 30, 2009, 07:22:28 pm »

$\cos^2 \left(\frac{x}{2} \right) = \frac{1}{2} \left( 1 + \cos(x) \right)$

And then you can sub in $\frac{\pi}{4}$ in there.

[Be careful what answer you pick, make sure you read the question to decide which quadrant it's in]

sachinmachin · « **Reply #7 on:** September 30, 2009, 07:24:35 pm »

Quote from: TrueTears on September 30, 2009, 07:22:28 pm

$\cos^2 \left(\frac{x}{2} \right) = \frac{1}{2} \left( 1 + \cos(x) \right)$

And then you can sub in $\frac{\pi}{4}$ in there.

[Be careful what answer you pick, make sure you read the question to decide which quadrant it's in]

makes so much more sense now. thankyou TT

dejan91 · « **Reply #8 on:** September 30, 2009, 07:28:44 pm »

TrueTears, I swear.... Stop being so quick on the posts! Was posting the exact same thing but am just 10x slower on latex.....

I'm gonna post anyway so my time doen't go to waste

:

$cos (2x) = 2 cos^2 (x)-1$
$cos(x) = \sqrt {\frac{1}{2}(cos((2x) + 1)}$

And use $x = \frac {\pi}{8}$ .

TrueTears · « **Reply #9 on:** September 30, 2009, 07:31:34 pm »

Quote from: dejan91 on September 30, 2009, 07:28:44 pm

TrueTears, I swear.... Stop being so quick on the posts! Was posting the exact same thing but am just 10x slower on latex.....

I'm gonna post anyway so my time doen't go to waste :

$cos (2x) = 2 cos^2 (x)-1$
$cos(x) = \sqrt {\frac{1}{2}(cos((2x) + 1)}$

And use $x = \frac {\pi}{8}$ .

hahahahahaha sorry!

dejan91 · « **Reply #10 on:** September 30, 2009, 07:34:15 pm »

Quote from: TrueTears on September 30, 2009, 07:31:34 pm

Quote from: dejan91 on September 30, 2009, 07:28:44 pm
TrueTears, I swear.... Stop being so quick on the posts! Was posting the exact same thing but am just 10x slower on latex.....

I'm gonna post anyway so my time doen't go to waste :

$cos (2x) = 2 cos^2 (x)-1$
$cos(x) = \sqrt {\frac{1}{2}(cos((2x) + 1)}$

And use $x = \frac {\pi}{8}$ .

hahahahahaha sorry!

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