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## apple_pi 3 years ago Differentiate sin(x) from first principles

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1. hartnn

use sin A-sin B formula, do u know it ?

2. apple_pi

the sin(A+B)?

3. hartnn

nopes, sin A - sin B on sin(x+h) - sin x.

4. apple_pi

I was thinking on starting out like this: $\lim_{h \rightarrow 0}\frac{ \sin (x+h) - \sin(x) }{ h }$

5. hartnn

thats the correct start, i was giving a hint on how would u start simplifying numerator. using sin A-sin B with A=x+h,B=x

6. apple_pi

us, would you expand sin(x+h)?

7. apple_pi

*um

8. hartnn

i think that would also work, there u have to separate out denominator's h to all numerator terms.....i personally find it easier to do with sin A-sin B...and less lengthy

9. apple_pi

how does sinA-sinB work?

10. hartnn

sin A - sin B = 2 cos (A+B)/2 sin (A-B)/ 2 right ? so sin (x+h)-sin x = 2 cos (x+h/2) sin (h/2) ok ? now put h=0 directly in cos term and use sin t/t = 1 in sin term...

11. apple_pi

sorry, don't know sinA-sinB = 2 cos((A+B)/2)sin((A+B)/2)

12. Callisto

It's... actually $sinA - sinB = 2cos\frac{A+B}{2}sin\frac{A-B}{2}$

13. Skaematik

Q: Is that the product to sums formula?

14. hartnn

then u better go with sin(x+h) = sin x cos h + cos x sin h ....

15. Callisto

sum-to-product indeed.

16. apple_pi

lim... [ sin(x)cos(h)+cos(x)sin(h)-sin(x) ] / h lim... [ sin(x)(cos(h)-1)+cos(x)sin(h) ] / h then what?

17. hartnn

can u solve lim x->0 (1-cos x)/x = 1/2 ?? pretty standard. the other limit is pretty easy, just use sin h/h =1

18. Callisto

Hmmm... Isn't it lim x->0 (1-cos x)/x = 0?

19. apple_pi

isn't it lim h->0?

20. hartnn

made a mistake....wait.yup its 0,sorry.

21. hartnn

but basically u need to separate the numerator in 2 limits.

22. hartnn

so your 1st limit will be 0 as pointed out correctly and 2nd limit will be just cos x *1 = cos x

23. Callisto

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

24. Callisto

Perhaps one more step for you... $=sin(x)\lim_{h \rightarrow 0}\frac{(cos(h)-1)}{h} \ +cos(x) \lim_{h \rightarrow 0}\frac{sin(h) }{h}$ I think you can evaluate the limit now..

25. apple_pi

how are theses limits calculated?

26. Callisto

*limits

27. Callisto

As @hartnn mentioned, use these: $\lim_{x \rightarrow 0}\frac{1-cos(x)}{x} =0$$\lim_{x \rightarrow 0}\frac{sin(x)}{x} =1$

28. hartnn

actually the 1st one isn't standard, u need to solve it by multiplying and dividing by (1+cos x) which gives sin^2 x in numerator and using 2nd limit to get answer as 0

29. apple_pi

yeah um, how are these proven? (Sorry if I'm being a bit fussy)

30. Callisto

@apple_pi Do you have time to watch a video for explanations?

31. apple_pi

maybe later... just send me the links

32. hartnn

there are many ways to prove lim sin x/x =1 the simplest one is the use of L'Hospitals rule, but since u are starting limits, i guess u don't know about this rule..

33. Callisto
34. apple_pi

Oh yeah, MIT OCW is good. Does this show the whole d/dx sin(x)?

35. Callisto

Yup!

36. apple_pi

thanks...turns out I downloaded this ages ago, but stopped after lecture 2 and jumped about places

37. Callisto

Here it goes. The notes I've copied from that lecture. IMG.pdf should contain all three pages.

38. apple_pi

Thanks a heap. Can you show me how sinA-sinB = ...

39. Callisto

Do you know how to work out the product-to-sum formula?

40. apple_pi

whats that?

41. Callisto

Example: sinAcosB = (1/2) [ sin(A+B) +sin(A-B)]

42. hartnn

that u can get using sin (x-y) and sin (x+y) formula, u know that right ? just solve sin(x+y) - sin (x-y) = ....... and then put x+y = A x-y = B so x= (A+B)/2 y=(A-B)/2.

43. apple_pi

sorry, no

44. apple_pi

no @Callisto

45. apple_pi

yes @hartnn

46. Callisto

hartnn explained, I think :)

47. hartnn

so whats sin(x+y) - sin (x-y)= ?

48. apple_pi

sin(x)cos(y)+cos(x)sin(y)-sin(x)cos(y)+cos(x)sin(y) = 2cos(x)sin(y)

49. hartnn

thats correct, now just put values of x and y as i mentioned, u understood, how i found x and y in terms of A and b .. ?

50. apple_pi

um, no

51. hartnn

x+y=A x-y=B Adding x+x=A+B --->x=(A+B)/2 i think y can can find now...

52. apple_pi

(A+B)/2+y=A y=(A-B)/2

53. hartnn

yup, any more doubts?

54. apple_pi

uh, so does that mean sin(A)-sin(B)=2cos((A+B)/2)sin((A-B)/2)

55. apple_pi

@hartnn

56. hartnn

exactly, yes.

57. apple_pi

OK, thank you

58. hartnn

similarly u can find sin A+sin B cos A + cos B cos A - cos B and remember them as they are standard formulas.

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