sin (pi/3) = (square root of 3)/2atakach99 said:Given x is small, show that
sin (pie/3 + x) approximatly equal to 1/2(square root 3 + x)
Oh. That's smart!!! I better learn how to think outside the norm.tommykins said:I'd use limits as x -> 0 and sub x = 0 in.
sin 0.045 = 0.0449848140...atakach99 said:ok guys i know this is a sill question sorry
evaluate, correct to 3 decimal places
sin 0.045
when i type it into my calculator i get a weird answer can somebody show me how to do this question plz
I know few kids in my class who don't even know whether they have to use radian or degree mode. So, it's quite common.atakach99 said:i feel so embaressed right now
Lol I don't like limit theorems in uni maths either.atakach99 said:i feel so embaressed right now
You should take note of which question level it is, since this is 2unit - it only requires a simple method.lyounamu said:Oh. That's smart!!! I better learn how to think outside the norm.
L.H.S.=lim x -> 0 sin (pi/3 +x)
= sin (pi/3 +0)
= sin (pi/3)
= (square root of 3)/2
= 1/2 (square root of 3 + 0)
= 1/2 (square root of 3 + x) (since x = 0 in this case).
= R.H.S
Thanks to tommykins for great advice. I would not have solved it without the advice he gave me.
Don't you use this level of working out in Mathematics?tommykins said:You should take note of which question level it is, since this is 2unit - it only requires a simple method.
It really depends on how you think about it. Yeah this was in the prelim year course, it's mainly just simplify an expression and sub in the given limit. It gets harder as yuo go however.lyounamu said:Don't you use this level of working out in Mathematics?
I remember doing this kind of question back in Preliminary year.
Came back from dinner, looked at the question, now I feel my stomach churning my food clearly.tommykins said:It really depends on how you think about it. Yeah this was in the prelim year course, it's mainly just simplify an expression and sub in the given limit. It gets harder as yuo go however.
I had a question in my exam (3unit) where it was -
lim x -> pi/4 (sinx - cosx)/(x-pi/4)
In which I had no idea where to start, you couldn't sub in x = pi/4 since it makes the denominator 0 which then leads to it being undefined.
If you feel like having a go at it, post your answer here and I'll tell you if you're right or not
Hint - rewrite the numerator sinx - cosx in a different form.lyounamu said:Came back from dinner, looked at the question, now I feel my stomach churning my food clearly.
I don't get it actually. But let me try.
EDIT: I think the answer is 0 somehow.
I already did. I wrote that in t-formula. Did I have to write that in other form? (like sinx = 2(sinx/2)cos(x/2)?)tommykins said:Hint - rewrite the numerator sinx - cosx in a different form.
Lol, just do any method you want, although the solution I have are pretty ridiculous and I was "wtf?" when I looked at it.lyounamu said:I already did. I wrote that in t-formula. Did I have to write that in other form? (like sinx = 2(sinx/2)cos(x/2)?)
I somehow need to get rid of the bottom one (or change it) so that I can work it out. It's not as easy as it looks like. I bloody used the t-formula and it gets more complicated.....tommykins said:Lol, just do any method you want, although the solution I have are pretty ridiculous and I was "wtf?" when I looked at it.