anonymous
  • anonymous
Derive sinx using first principles
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
sin(x) = y/r .....
anonymous
  • anonymous
Wait May I show you how far I got?
anonymous
  • anonymous
go get em!

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amistre64
  • amistre64
you can
anonymous
  • anonymous
you need the addition angle forumla\[sin(x+h)-sin(x)=sin(x)cos(h)+cos(x)sin(h)\]
anonymous
  • anonymous
Ok, imagine this is the limit h -> 0 f(x + h) - f(x) / h sin(x + h) - sin(x) / h Applied compound angle formula:
amistre64
  • amistre64
ohhh... i dint know we could use the sin function itself lol
anonymous
  • anonymous
lim h-> 0 [(sinxcosh + cosxsinh - sinx) / h]
anonymous
  • anonymous
Now what do I do?
anonymous
  • anonymous
I'm not sure what we are deriving
amistre64
  • amistre64
factor out a sin
anonymous
  • anonymous
that 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
  • anonymous
ohhh derivative of sin(x) haha
anonymous
  • anonymous
so you need \[lim_{h->0}\frac{sin(h)}{h} = 1\]
anonymous
  • anonymous
and \[lim_{h->0}\frac{cos(h)}{h}=0\]
anonymous
  • anonymous
leaving just \[cos(X)\]
anonymous
  • anonymous
Oh really? cosh/h = 0?
anonymous
  • anonymous
I didn't know that
anonymous
  • anonymous
Ok wait so
anonymous
  • anonymous
it is \[lim_{x->0}\frac{cos(h)}{h}=0\]
anonymous
  • anonymous
you need to know this to prove your result. probably comes in the text right before this.
anonymous
  • anonymous
lim cos(h)/h as h goes to 0 doesnt exist
anonymous
  • anonymous
actually it does and it is 0
anonymous
  • anonymous
So so this is what I have now: lim h-> 0 [(sinx(cosh - 1) + sinhcosx) / h]
anonymous
  • anonymous
lets start again.
anonymous
  • anonymous
how, from the left it approaches -infinity from the right it approaches +infinity
anonymous
  • anonymous
the 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
  • anonymous
oops sorry rsvitale is right and i am wrong
anonymous
  • anonymous
my mistake i apologize
anonymous
  • anonymous
(cosh-1)/h goes to 0 as h goes to 0 though
anonymous
  • anonymous
you need two limits: \[lim_{h->0}\frac{sin(h)}{h}=1\] and \[lim_{h->0}\frac{cos(h)-1}{h}=0\]
anonymous
  • anonymous
yes you are right. my fault
anonymous
  • anonymous
Sorry but I am getting a little confused as to what is right and what is wrong here.. :\
anonymous
  • anonymous
i am wrong. but if you like i will write out the correct version since i messed up
anonymous
  • anonymous
hey 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
  • anonymous
\[lim_{h->0}\frac{sin(x+h)-sin(x)}{h}\] \[lim_{h->0}\frac{sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}\] \[=lim_{h->0}\frac{sin(x)(cos(h)-1) + cos(x)sin(x)}{h}\]
anonymous
  • anonymous
break into two parts. \[sin(x)lim_{h->0}\frac{cos(h)-1}{h} + cos(x)lim_{h->0}\frac{sin(h)}{h}\]
anonymous
  • anonymous
the first limit is 0, the second limit is 1 leaving\[cos(x)\]
anonymous
  • anonymous
satellite73, you wrote sinx(cosh - 1) + cosxsinx <-- isn't it suppose to be sinhcosx?

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