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