anonymous
  • anonymous
how do I find sin 5ᴨ/8, cos 5ᴨ/8 and tan 5ᴨ/8 using trig identities
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
What is ᴨ
anonymous
  • anonymous
use formula for sin4x
anonymous
  • anonymous
the problem doesn't say...just says to use trigonometric identities to find the values

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anonymous
  • anonymous
But what is "ᴨ" is it theta or something else. What is it?
anonymous
  • anonymous
expand sin4x, put x as 5pi/8 and equate the expa nsion of sin4x to 1
anonymous
  • anonymous
Is it pi?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Sin ((270-45)/2)=sin (5(pi)/8)
anonymous
  • anonymous
Do you understand this step?
anonymous
  • anonymous
Do you understand the step "Sin ((270-45)/2)=sin (5(pi)/8)"
anonymous
  • anonymous
not really, but i think you got the 270 from 3pi/2 and the 45 from pi/4....
anonymous
  • anonymous
Well, I am trying to express the thing in degree format. (Sorry for using jumbled form). By the term I mean 180 degree
anonymous
  • anonymous
By the term pi mean 180 degree
anonymous
  • anonymous
\[Sin ((3\pi/2-\pi/4)/2)=sin (5\pi/8)\]
anonymous
  • anonymous
Now does that makes sense?
anonymous
  • anonymous
yes
anonymous
  • anonymous
*make
anonymous
  • anonymous
Now can you do the rest, or I will have to show you every steps.
anonymous
  • anonymous
so, sin (225/2)=sin (112.5)? you might have to show me the steps.
anonymous
  • anonymous
or is it sin (5pi/4 divided by 2)
anonymous
  • anonymous
Yes, thats true, but thats not helpful here. \[Sin ((3\pi/2-\pi/4)/2)=sin (5\pi/8)\] \[=>sin (5\pi/8)=\sqrt(0.5(1-cos(3\pi/2-\pi/4))\]
anonymous
  • anonymous
Now you got to use addition formula for cos to solve the part with cos, and then it will be done
anonymous
  • anonymous
I am totally lost .....this does not look at all familiar.....
anonymous
  • anonymous
are you familiar with double angle formula for trig?
anonymous
  • anonymous
i'm sure we went over them....but nothing coming to mind
anonymous
  • anonymous
Then you can't expect to get this thing within your understandability range
anonymous
  • anonymous
that's why I am having trouble....thought if someone walked me through it, it would make sense
anonymous
  • anonymous
If you wish I can tutor you (in exchange of some nominal payment) :)
anonymous
  • anonymous
Seems like you are not interested. So may I leave?
anonymous
  • anonymous
didn't know i had to pay for help
anonymous
  • anonymous
No no not that
anonymous
  • anonymous
you can get free help here for ever
anonymous
  • anonymous
But I don't think they will be very helpful
anonymous
  • anonymous
So incase you need personal help
anonymous
  • anonymous
I mean through audio chatting and virtual black board
anonymous
  • anonymous
I can't teach you everything here. This place is too inconvenient to type even a simple equation
anonymous
  • anonymous
I mean no one can
anonymous
  • anonymous
thanks for the offer, but rather just stay here
anonymous
  • anonymous
Thats great. No problem

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