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Differentiate sin(x) from first principles

Mathematics
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use sin A-sin B formula, do u know it ?
the sin(A+B)?
nopes, sin A - sin B on sin(x+h) - sin x.

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Other answers:

I was thinking on starting out like this: \[\lim_{h \rightarrow 0}\frac{ \sin (x+h) - \sin(x) }{ h }\]
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
us, would you expand sin(x+h)?
*um
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
how does sinA-sinB work?
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...
sorry, don't know sinA-sinB = 2 cos((A+B)/2)sin((A+B)/2)
It's... actually \[sinA - sinB = 2cos\frac{A+B}{2}sin\frac{A-B}{2}\]
Q: Is that the product to sums formula?
then u better go with sin(x+h) = sin x cos h + cos x sin h ....
sum-to-product indeed.
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?
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
Hmmm... Isn't it lim x->0 (1-cos x)/x = 0?
isn't it lim h->0?
made a mistake....wait.yup its 0,sorry.
but basically u need to separate the numerator in 2 limits.
so your 1st limit will be 0 as pointed out correctly and 2nd limit will be just cos x *1 = cos x
\[\lim_{h \rightarrow 0}\frac{sin(x)(cos(h)-1)+cos(x)sin(h) }{h}\]\[=\lim_{h \rightarrow 0}(\frac{sin(x)(cos(h)-1)}{h}\ \ +\frac{cos(x)sin(h) }{h})\]\[=\lim_{h \rightarrow 0}sin(x)\frac{(cos(h)-1)}{h} \ +\lim_{h \rightarrow 0}cos(x)\frac{sin(h) }{h}\]\[=...\]
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..
how are theses limits calculated?
*limits
As @hartnn mentioned, use these: \[\lim_{x \rightarrow 0}\frac{1-cos(x)}{x} =0\]\[\lim_{x \rightarrow 0}\frac{sin(x)}{x} =1\]
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
yeah um, how are these proven? (Sorry if I'm being a bit fussy)
@apple_pi Do you have time to watch a video for explanations?
maybe later... just send me the links
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..
http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/lecture-3-derivatives/ From 3:05.
Oh yeah, MIT OCW is good. Does this show the whole d/dx sin(x)?
Yup!
thanks...turns out I downloaded this ages ago, but stopped after lecture 2 and jumped about places
Here it goes. The notes I've copied from that lecture. IMG.pdf should contain all three pages.
Thanks a heap. Can you show me how sinA-sinB = ...
Do you know how to work out the product-to-sum formula?
whats that?
Example: sinAcosB = (1/2) [ sin(A+B) +sin(A-B)]
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.
sorry, no
hartnn explained, I think :)
so whats sin(x+y) - sin (x-y)= ?
sin(x)cos(y)+cos(x)sin(y)-sin(x)cos(y)+cos(x)sin(y) = 2cos(x)sin(y)
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 .. ?
um, no
x+y=A x-y=B Adding x+x=A+B --->x=(A+B)/2 i think y can can find now...
(A+B)/2+y=A y=(A-B)/2
yup, any more doubts?
uh, so does that mean sin(A)-sin(B)=2cos((A+B)/2)sin((A-B)/2)
exactly, yes.
OK, thank you
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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