Question

Differentiate sin(x) from first principles

Answer 1

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

Answer 2

the sin(A+B)?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

*um

Answer 8

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

Answer 9

how does sinA-sinB work?

Answer 10

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...

Answer 11

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

Answer 12

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

Answer 13

Q: Is that the product to sums formula?

Answer 14

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

Answer 15

sum-to-product indeed.

Answer 16

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?

Answer 17

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

Answer 18

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

Answer 19

isn't it lim h->0?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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..

Answer 25

how are theses limits calculated?

Answer 26

*limits

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

maybe later... just send me the links

Answer 32

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..

Answer 33

http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/lecture-3-derivatives/
From 3:05.

Answer 34

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

Answer 35

Yup!

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

whats that?

Answer 41

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

Answer 42

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.

Answer 43

sorry, no

Answer 44

no @Callisto

Answer 45

yes @hartnn

Answer 46

hartnn explained, I think :)

Answer 47

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

Answer 48

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

Answer 49

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 .. ?

Answer 50

um, no

Answer 51

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

Answer 52

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

Answer 53

yup, any more doubts?

Answer 54

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

Answer 55

@hartnn

Answer 56

exactly, yes.

Answer 57

OK, thank you

Answer 58

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