Derive sinx using first principles

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Derive sinx using first principles

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

sin(x) = y/r .....
Wait May I show you how far I got?
go get em!

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question