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hartnn Group TitleBest ResponseYou've already chosen the best response.2
use sin Asin B formula, do u know it ?
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
the sin(A+B)?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
nopes, sin A  sin B on sin(x+h)  sin x.
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
I was thinking on starting out like this: \[\lim_{h \rightarrow 0}\frac{ \sin (x+h)  \sin(x) }{ h }\]
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
thats the correct start, i was giving a hint on how would u start simplifying numerator. using sin Asin B with A=x+h,B=x
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
us, would you expand sin(x+h)?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
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 Asin B...and less lengthy
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
how does sinAsinB work?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
sin A  sin B = 2 cos (A+B)/2 sin (AB)/ 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...
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
sorry, don't know sinAsinB = 2 cos((A+B)/2)sin((A+B)/2)
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
It's... actually \[sinA  sinB = 2cos\frac{A+B}{2}sin\frac{AB}{2}\]
 2 years ago

Skaematik Group TitleBest ResponseYou've already chosen the best response.0
Q: Is that the product to sums formula?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
then u better go with sin(x+h) = sin x cos h + cos x sin h ....
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
sumtoproduct indeed.
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
can u solve lim x>0 (1cos x)/x = 1/2 ?? pretty standard. the other limit is pretty easy, just use sin h/h =1
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
Hmmm... Isn't it lim x>0 (1cos x)/x = 0?
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
isn't it lim h>0?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
made a mistake....wait.yup its 0,sorry.
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
but basically u need to separate the numerator in 2 limits.
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
so your 1st limit will be 0 as pointed out correctly and 2nd limit will be just cos x *1 = cos x
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
\[\lim_{h \rightarrow 0}\frac{sin(x)(cos(h)1)+cos(x)sin(h) }{h}\]\[=\lim_{h \rightarrow 0}(\frac{sin(x)(cos(h)1)}{h}\ \ +\frac{cos(x)sin(h) }{h})\]\[=\lim_{h \rightarrow 0}sin(x)\frac{(cos(h)1)}{h} \ +\lim_{h \rightarrow 0}cos(x)\frac{sin(h) }{h}\]\[=...\]
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
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..
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
how are theses limits calculated?
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
As @hartnn mentioned, use these: \[\lim_{x \rightarrow 0}\frac{1cos(x)}{x} =0\]\[\lim_{x \rightarrow 0}\frac{sin(x)}{x} =1\]
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
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
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
yeah um, how are these proven? (Sorry if I'm being a bit fussy)
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
@apple_pi Do you have time to watch a video for explanations?
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
maybe later... just send me the links
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
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..
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
Oh yeah, MIT OCW is good. Does this show the whole d/dx sin(x)?
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
thanks...turns out I downloaded this ages ago, but stopped after lecture 2 and jumped about places
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
Here it goes. The notes I've copied from that lecture. IMG.pdf should contain all three pages.
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
Thanks a heap. Can you show me how sinAsinB = ...
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
Do you know how to work out the producttosum formula?
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
whats that?
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
Example: sinAcosB = (1/2) [ sin(A+B) +sin(AB)]
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
that u can get using sin (xy) and sin (x+y) formula, u know that right ? just solve sin(x+y)  sin (xy) = ....... and then put x+y = A xy = B so x= (A+B)/2 y=(AB)/2.
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
no @Callisto
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
yes @hartnn
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
hartnn explained, I think :)
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
so whats sin(x+y)  sin (xy)= ?
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
sin(x)cos(y)+cos(x)sin(y)sin(x)cos(y)+cos(x)sin(y) = 2cos(x)sin(y)
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
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 .. ?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
x+y=A xy=B Adding x+x=A+B >x=(A+B)/2 i think y can can find now...
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
(A+B)/2+y=A y=(AB)/2
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
yup, any more doubts?
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
uh, so does that mean sin(A)sin(B)=2cos((A+B)/2)sin((AB)/2)
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
exactly, yes.
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
OK, thank you
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
similarly u can find sin A+sin B cos A + cos B cos A  cos B and remember them as they are standard formulas.
 2 years ago
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