Ask your own question, for FREE!
Mathematics 10 Online
OpenStudy (anonymous):

use a double- angle identity to find the exact value sin22.5^ocos22.5^o

OpenStudy (anonymous):

why I am still at 19 fans, after spending the last 5 hours working my *** out?

OpenStudy (anonymous):

if you help me with this i will become a fan

OpenStudy (anonymous):

ok, one min

OpenStudy (anonymous):

the double angle identities are as follows: \[\cos (A - B) = (\cos A)(\cos B) + (\sin A)(\sin B) \] \[\cos (A + B) = (\cos A)\cos B - (\sin A)\sin B \] \[\sin (A + B) = (\sin A)(\cos B) + (\cos A)(\sin B) \] \[\sin (A - B) = (\sin A)(\cos B) - (\cos A)(\sin B) \] alright andy :) you can lead from here ^_^

OpenStudy (anonymous):

hmm, am I mixing it all up with the double angle identities? those are the sin and cos indentities , hmm, please correct me if I'm wrong ^_^

OpenStudy (anonymous):

double angle identities go like this, egzample: sin(2x)=2sinxcosx

OpenStudy (nowhereman):

That is just the above with setting A=B

OpenStudy (anonymous):

no the identities are sin2A=2sineAcosA cos2A-cos^2A-sin^2A =1-2sin^2A =2cosA-1

OpenStudy (anonymous):

or right lol , thank you, alright angoo, give it a try ^_^

OpenStudy (anonymous):

ok I am bad at this, I guess I must have skipped school the day this stuff was tought

OpenStudy (anonymous):

give it a try :)

OpenStudy (anonymous):

yeah same here

OpenStudy (anonymous):

nah, stary, you go ahead, I dont get this stuff yet

OpenStudy (anonymous):

ask your beloved google lol

OpenStudy (anonymous):

well, Google can give me examples, and explanations, and I can learn from them, but it would take too much time... so just go ahead and do it mimi

OpenStudy (anonymous):

this is my new signature btw: ▓▒░╔ᴧᴨᴅᴙiᴜs╖░▒▓

OpenStudy (anonymous):

LOL

OpenStudy (anonymous):

alright LoveL , let me get this clear : you want to find the answer of : sin(22.5)cos(22.5)? or do you want to find the exact angle for sin(22.5)=? or cos(22.5)=? ^_^ lol, it's filled with mystery, but I like it :)

OpenStudy (nowhereman):

I will show you, first of all: \[\sin (22.5^\circ)\cos(22.5^\circ) = \sin(\frac{180^\circ}{8})\cos(\frac{180^\circ}{8}) = \sin(\frac{π}{8})\cos(\frac π 8)\]

OpenStudy (anonymous):

nvm, follow up with nowhereman ^_^ he'll lead you through.

OpenStudy (anonymous):

so, loveinglife, do you feel happy today? because I do, but I could become even more happy if you gift me a fan =D In that case, I will give you a fan too, so you become a lifesaver, and I become a superstar, a clear winwin =D

OpenStudy (anonymous):

LOL, and you're still on it

OpenStudy (anonymous):

@BecomeMyFan im your fan now. thank you for trying

OpenStudy (nowhereman):

Then \[\sin(\frac π 8)\cos(\frac π 8) = \frac 1 2 \cdot \left(2\sin(\frac π 8)\cos(\frac π 8)\right)\] Thus by applying that double arc formula on the inner part (in reverse direction if you want) you get \[=\frac 1 2 \sin(2\frac π 8) = \frac 1 2\sin(\frac π 4) = \frac 1 4 \sqrt 2\]

OpenStudy (anonymous):

Oh my god your awesome. thank you soo much you just saved my trig grade

OpenStudy (nowhereman):

You're welcome :-D

OpenStudy (anonymous):

love, are you in highschool?

OpenStudy (anonymous):

or in college

OpenStudy (anonymous):

?

OpenStudy (anonymous):

because if you are doing this stuff in high school, I must have skipped ALOT of it

OpenStudy (anonymous):

im saddly in college and i thought i was good in math till i got here

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

why am I still a mear star?

OpenStudy (anonymous):

I wanna be a superstar

OpenStudy (anonymous):

it was supposed to rhyme

OpenStudy (anonymous):

=D

OpenStudy (anonymous):

sorry cant help you there but im done, but i will probably be back at some point

OpenStudy (anonymous):

it is part of a poem I am writting about OpenStudy

OpenStudy (anonymous):

=)

OpenStudy (anonymous):

you write poems?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!