## anonymous 5 years ago how do I find sin 5ᴨ/8, cos 5ᴨ/8 and tan 5ᴨ/8 using trig identities

1. anonymous

What is ᴨ

2. anonymous

use formula for sin4x

3. anonymous

the problem doesn't say...just says to use trigonometric identities to find the values

4. anonymous

But what is "ᴨ" is it theta or something else. What is it?

5. anonymous

expand sin4x, put x as 5pi/8 and equate the expa nsion of sin4x to 1

6. anonymous

Is it pi?

7. anonymous

yes

8. anonymous

Sin ((270-45)/2)=sin (5(pi)/8)

9. anonymous

Do you understand this step?

10. anonymous

Do you understand the step "Sin ((270-45)/2)=sin (5(pi)/8)"

11. anonymous

not really, but i think you got the 270 from 3pi/2 and the 45 from pi/4....

12. anonymous

Well, I am trying to express the thing in degree format. (Sorry for using jumbled form). By the term I mean 180 degree

13. anonymous

By the term pi mean 180 degree

14. anonymous

$Sin ((3\pi/2-\pi/4)/2)=sin (5\pi/8)$

15. anonymous

Now does that makes sense?

16. anonymous

yes

17. anonymous

*make

18. anonymous

Now can you do the rest, or I will have to show you every steps.

19. anonymous

so, sin (225/2)=sin (112.5)? you might have to show me the steps.

20. anonymous

or is it sin (5pi/4 divided by 2)

21. anonymous

Yes, thats true, but thats not helpful here. $Sin ((3\pi/2-\pi/4)/2)=sin (5\pi/8)$ $=>sin (5\pi/8)=\sqrt(0.5(1-cos(3\pi/2-\pi/4))$

22. anonymous

Now you got to use addition formula for cos to solve the part with cos, and then it will be done

23. anonymous

I am totally lost .....this does not look at all familiar.....

24. anonymous

are you familiar with double angle formula for trig?

25. anonymous

i'm sure we went over them....but nothing coming to mind

26. anonymous

Then you can't expect to get this thing within your understandability range

27. anonymous

that's why I am having trouble....thought if someone walked me through it, it would make sense

28. anonymous

If you wish I can tutor you (in exchange of some nominal payment) :)

29. anonymous

Seems like you are not interested. So may I leave?

30. anonymous

didn't know i had to pay for help

31. anonymous

No no not that

32. anonymous

you can get free help here for ever

33. anonymous

But I don't think they will be very helpful

34. anonymous

So incase you need personal help

35. anonymous

I mean through audio chatting and virtual black board

36. anonymous

I can't teach you everything here. This place is too inconvenient to type even a simple equation

37. anonymous

I mean no one can

38. anonymous

thanks for the offer, but rather just stay here

39. anonymous

Thats great. No problem