Question

sin(3) =

Answer 1

f(x)=sin(x)
3 is close to pi
f(pi)=sin(pi)=0
f'(x)=cos(x)
f'(pi)=cos(pi)=-1
so we have the linear approximate
\[f(x) \approx -(x-\pi)+0 \\ f(3) \approx -(3-\pi)=\pi-3 \approx 3.14-3=.14\]

Answer 2

but you probably aren't looking for something like that

Answer 3

i'm looking for an exact form

Answer 4

(your answer; @freckles, is accurate to 2 decimal places )

Answer 5

For a hint:

3 = 1 + 2

Answer 6

actually 3 if I use the 3.141 approximation instead of the 3.14 for pi :p
but yeah it is definitely not an exact answer either way

Answer 7

so I'm guess we can find sin(1) and cos(1) and sin(2) and cos(2) exactly
and use the sum identity for sin(3)=sin(2+1)

Answer 8

YES, the sum identity,
[final form will have a sin(1) and a cos(2) in it]

Answer 9

\[\sin(2+1)=\sin(2)\cos(1)+\sin(1)\cos(2)\]
hmmm so you are saying we can either
evaluate sin(2) and cos(1) but not sin(1) and cos(2) which seems unlikely to me because if we can evaluate sin(2) and cos(1) then we should be able to evaluate sin(1) and cos(2)


so what I think is you want me to write it just in terms of sin(1) and cos(2)

Answer 10

well 2 and 1 > pi
so we can use pythagorean identity
\[\sin(3)=\sqrt{1-(\cos(2))^2} \sqrt{1-(\sin(1))^2}+\sin(1)\cos(2)\]

Answer 11

oops I mean 2 and 1 <pi

Answer 12

\[\sin(2)=\sin(1+1)=\sin(1)\cos(1)+\cos(1)\sin(1)=2\sin(1)\cos(1)\]

Answer 13

here is what I'm trying to get out
1<pi/2
so cos(1) >0

and
pi/2<2<pi
so sin(2)>0

just trying to say why I chose the positives and not the negatives for by Pythagorean identity solving junk

Answer 14

\(\sin(3x) = 3\sin x-4\sin^3x\)

\(\sin(3*1) = 3\sin 1 - 4\sin^31\)

Answer 15

oh so you wanted this:
\[\sin(2+1)=\sin(2)\cos(1)+\sin(1)\cos(2) \\ =2 \sin(1) \cos(1) \cos(1)+\sin(1)\cos(2) \\ =2 \sin(1) \cos^2(1)+\sin(1)\cos(2) \]\
\[=2\sin(1)(1-\sin^2(1))+\sin(1)\cos(2) \\ 2\sin(1)-2\sin^3(1)+\sin(1)\cos(2)\]
so this is what you want ?
this is in terms of sin(1) and cos(2) ?

Answer 16

What is so specially about \(\sin 1\) or \(\sin 2\)?

Answer 17

1 and 2 are simpler than 3

Answer 18

I must leave but I will check back on this later :)

Answer 19

\[\sin(3)\\=\sin(1+2)\\
=\sin(1)\cos(2)+\cos(1)\sin(2)\\
=\sin(1)\cos(2)+\cos(1)[2\sin 1\cos 1]\\
=\sin(1)\cos(2)+2\sin(1)\cos^2(1)\\
=\sin(1)\left(\cos(2)+2[\tfrac12(1+\cos(2))]\right)\\
=\sin(1)(2\cos(2)+1)\]

Answer 20

this form looks easy for approximating sin(3) by hand \[\sin(3)=\dfrac{1}{7+\dfrac{1}{11+\dfrac{1}{1+\dfrac{1}{1+\cdots}}}}\]

Answer 21

How did you get that continued fraction @rational ?

Answer 22

Unkle please see robjohn's reply in this thread
http://math.stackexchange.com/questions/298029/how-can-we-convert-sin-function-into-continued-fraction

Answer 23

i don't fully understand, but as you can see simply plugging in x = 3 in the final result gives us a continued fraction for our specific case

Answer 24

https://en.wikipedia.org/wiki/Sine#Continued_fraction

Answer 25

Hmm, so i get:

(i think robjohn has a typo in his final result (4))

\[\sin\color{green}3=\dfrac {\color{green}3}{1+\dfrac{\color{green}3^2}{2\cdot3-\color{green}3^{2}+\dfrac{6\cdot\color{green}3^2}{4\cdot5-\color{green}3^2+\dfrac{4\cdot5\cdot\color{green}3^2}{6\cdot7-\color{green}3^2+_{\displaystyle{^\ddots ~ 2n(2n+1)-\color{green}3^2+\dfrac{2n(2n+1)\color{green}3^2}{P_n(\color{green}3)/P_{n+1}(\color{green}3)}}}}}}}\]

\[\qquad=\dfrac {3}{1+\dfrac9{6-9+\dfrac{72}{20-9+\dfrac{180}{42-9+_{\displaystyle{^\ddots ~}}}}}}\]

\[\qquad=\dfrac {3}{1+\dfrac9{-3+\dfrac{72}{11+\dfrac{180}{33+_{{^\ddots ~}}}}}}\]

Answer 26

How can we cancel common factors, or rearrange somehow to show our results are equal, (assuming they are)