Question

Derive sinx using first principles

Answer 1

sin(x) = y/r .....

Answer 2

Wait
May I show you how far I got?

Answer 3

go get em!

Answer 4

you can

Answer 5

you need the addition angle forumla\[sin(x+h)-sin(x)=sin(x)cos(h)+cos(x)sin(h)\]

Answer 6

Ok, imagine this is the limit h -> 0
f(x + h) - f(x) / h
sin(x + h) - sin(x) / h
Applied compound angle formula:

Answer 7

ohhh... i dint know we could use the sin function itself lol

Answer 8

lim h-> 0 [(sinxcosh + cosxsinh - sinx) / h]

Answer 9

Now what do I do?

Answer 10

I'm not sure what we are deriving

Answer 11

factor out a sin

Answer 12

that is your numerator. your denominator is just h.
you get
\[\frac{sin(x)cos(h)+cos(x)sin(h)}{h}=\frac{sin(x)cos(h)}{h} + \frac{cos(x)sin(h)}{h}\]

Answer 13

ohhh derivative of sin(x) haha

Answer 14

so you need \[lim_{h->0}\frac{sin(h)}{h} = 1\]

Answer 15

and \[lim_{h->0}\frac{cos(h)}{h}=0\]

Answer 16

leaving just \[cos(X)\]

Answer 17

Oh really? cosh/h = 0?

Answer 18

I didn't know that

Answer 19

Ok wait so

Answer 20

it is \[lim_{x->0}\frac{cos(h)}{h}=0\]

Answer 21

you need to know this to prove your result. probably comes in the text right before this.

Answer 22

lim cos(h)/h as h goes to 0 doesnt exist

Answer 23

actually it does and it is 0

Answer 24

So so this is what I have now:
lim h-> 0
[(sinx(cosh - 1) + sinhcosx) / h]

Answer 25

lets start again.

Answer 26

how, from the left it approaches -infinity from the right it approaches +infinity

Answer 27

the numerator is \[sin(x+h)-sin(x)=sin(x)cos(h) +sin(h)cos(x) - sin(x)=sin(x)(cos(h)-1)+sin(h)cos(x)\]

Answer 28

oops sorry rsvitale is right and i am wrong

Answer 29

my mistake i apologize

Answer 30

(cosh-1)/h goes to 0 as h goes to 0 though

Answer 31

you need two limits:
\[lim_{h->0}\frac{sin(h)}{h}=1\] and \[lim_{h->0}\frac{cos(h)-1}{h}=0\]

Answer 32

yes you are right. my fault

Answer 33

Sorry but I am getting a little confused as to what is right and what is wrong here.. :\

Answer 34

i am wrong. but if you like i will write out the correct version since i messed up

Answer 35

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

Answer 36

\[lim_{h->0}\frac{sin(x+h)-sin(x)}{h}\]
\[lim_{h->0}\frac{sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}\]
\[=lim_{h->0}\frac{sin(x)(cos(h)-1) + cos(x)sin(x)}{h}\]

Answer 37

break into two parts.
\[sin(x)lim_{h->0}\frac{cos(h)-1}{h} + cos(x)lim_{h->0}\frac{sin(h)}{h}\]

Answer 38

the first limit is 0, the second limit is 1 leaving\[cos(x)\]

Answer 39

satellite73, you wrote sinx(cosh - 1) + cosxsinx <-- isn't it suppose to be sinhcosx?