## anonymous 5 years ago Derive sinx using first principles

1. amistre64

sin(x) = y/r .....

2. anonymous

Wait May I show you how far I got?

3. anonymous

go get em!

4. amistre64

you can

5. anonymous

you need the addition angle forumla$sin(x+h)-sin(x)=sin(x)cos(h)+cos(x)sin(h)$

6. anonymous

Ok, imagine this is the limit h -> 0 f(x + h) - f(x) / h sin(x + h) - sin(x) / h Applied compound angle formula:

7. amistre64

ohhh... i dint know we could use the sin function itself lol

8. anonymous

lim h-> 0 [(sinxcosh + cosxsinh - sinx) / h]

9. anonymous

Now what do I do?

10. anonymous

I'm not sure what we are deriving

11. amistre64

factor out a sin

12. anonymous

that is your numerator. your denominator is just h. you get $\frac{sin(x)cos(h)+cos(x)sin(h)}{h}=\frac{sin(x)cos(h)}{h} + \frac{cos(x)sin(h)}{h}$

13. anonymous

ohhh derivative of sin(x) haha

14. anonymous

so you need $lim_{h->0}\frac{sin(h)}{h} = 1$

15. anonymous

and $lim_{h->0}\frac{cos(h)}{h}=0$

16. anonymous

leaving just $cos(X)$

17. anonymous

Oh really? cosh/h = 0?

18. anonymous

I didn't know that

19. anonymous

Ok wait so

20. anonymous

it is $lim_{x->0}\frac{cos(h)}{h}=0$

21. anonymous

you need to know this to prove your result. probably comes in the text right before this.

22. anonymous

lim cos(h)/h as h goes to 0 doesnt exist

23. anonymous

actually it does and it is 0

24. anonymous

So so this is what I have now: lim h-> 0 [(sinx(cosh - 1) + sinhcosx) / h]

25. anonymous

lets start again.

26. anonymous

how, from the left it approaches -infinity from the right it approaches +infinity

27. anonymous

the numerator is $sin(x+h)-sin(x)=sin(x)cos(h) +sin(h)cos(x) - sin(x)=sin(x)(cos(h)-1)+sin(h)cos(x)$

28. anonymous

oops sorry rsvitale is right and i am wrong

29. anonymous

my mistake i apologize

30. anonymous

(cosh-1)/h goes to 0 as h goes to 0 though

31. anonymous

you need two limits: $lim_{h->0}\frac{sin(h)}{h}=1$ and $lim_{h->0}\frac{cos(h)-1}{h}=0$

32. anonymous

yes you are right. my fault

33. anonymous

Sorry but I am getting a little confused as to what is right and what is wrong here.. :\

34. anonymous

i am wrong. but if you like i will write out the correct version since i messed up

35. anonymous

hey I found the derivation online and it's pretty good if you just want to look at that. http://oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/sine.html

36. anonymous

$lim_{h->0}\frac{sin(x+h)-sin(x)}{h}$ $lim_{h->0}\frac{sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}$ $=lim_{h->0}\frac{sin(x)(cos(h)-1) + cos(x)sin(x)}{h}$

37. anonymous

break into two parts. $sin(x)lim_{h->0}\frac{cos(h)-1}{h} + cos(x)lim_{h->0}\frac{sin(h)}{h}$

38. anonymous

the first limit is 0, the second limit is 1 leaving$cos(x)$

39. anonymous

satellite73, you wrote sinx(cosh - 1) + cosxsinx <-- isn't it suppose to be sinhcosx?