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anonymous
 5 years ago
Derive sinx using first principles
anonymous
 5 years ago
Derive sinx using first principles

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wait May I show you how far I got?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need the addition angle forumla\[sin(x+h)sin(x)=sin(x)cos(h)+cos(x)sin(h)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, imagine this is the limit h > 0 f(x + h)  f(x) / h sin(x + h)  sin(x) / h Applied compound angle formula:

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ohhh... i dint know we could use the sin function itself lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lim h> 0 [(sinxcosh + cosxsinh  sinx) / h]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not sure what we are deriving

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is your numerator. your denominator is just h. you get \[\frac{sin(x)cos(h)+cos(x)sin(h)}{h}=\frac{sin(x)cos(h)}{h} + \frac{cos(x)sin(h)}{h}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhh derivative of sin(x) haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you need \[lim_{h>0}\frac{sin(h)}{h} = 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and \[lim_{h>0}\frac{cos(h)}{h}=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0leaving just \[cos(X)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh really? cosh/h = 0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is \[lim_{x>0}\frac{cos(h)}{h}=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need to know this to prove your result. probably comes in the text right before this.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lim cos(h)/h as h goes to 0 doesnt exist

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually it does and it is 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So so this is what I have now: lim h> 0 [(sinx(cosh  1) + sinhcosx) / h]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how, from the left it approaches infinity from the right it approaches +infinity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the numerator is \[sin(x+h)sin(x)=sin(x)cos(h) +sin(h)cos(x)  sin(x)=sin(x)(cos(h)1)+sin(h)cos(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops sorry rsvitale is right and i am wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my mistake i apologize

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(cosh1)/h goes to 0 as h goes to 0 though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need two limits: \[lim_{h>0}\frac{sin(h)}{h}=1\] and \[lim_{h>0}\frac{cos(h)1}{h}=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes you are right. my fault

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry but I am getting a little confused as to what is right and what is wrong here.. :\

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am wrong. but if you like i will write out the correct version since i messed up

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey I found the derivation online and it's pretty good if you just want to look at that. http://oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/sine.html

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[lim_{h>0}\frac{sin(x+h)sin(x)}{h}\] \[lim_{h>0}\frac{sin(x)cos(h)+cos(x)sin(h)sin(x)}{h}\] \[=lim_{h>0}\frac{sin(x)(cos(h)1) + cos(x)sin(x)}{h}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0break into two parts. \[sin(x)lim_{h>0}\frac{cos(h)1}{h} + cos(x)lim_{h>0}\frac{sin(h)}{h}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the first limit is 0, the second limit is 1 leaving\[cos(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0satellite73, you wrote sinx(cosh  1) + cosxsinx < isn't it suppose to be sinhcosx?
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