|
|
|
Could you use a graphical justification? Just look at the graphs of the sine and cosine functions and it's apparent
I could show any given 2 angles that add to 90 on the right triangle and label them x and (pi/2-x) but wtf would the point of that be?
That allows you to derive the identity that you're looking for immediately so there's a very good reason to do that
|
okay so then I guess the problem must really be that simple
|
reread it and im confused too.
but yeah generally graphs aren't sufficient for showing that two things are identical
looks like the person below me explained it better than i could.
|
I think the proof would be:
2 (cos x) [cos (pi/2-x)] = 2(cos x)[(cos pi/2)(cos x) + (sin pi/2)(sin x)] <=breaking it out using angle sum identity
2(cos x)[(cos pi/2)(cos x) + (sin pi/2)(sin x)] = 2(cos x)[(0)(cos x) + (1)(sin x)]
2(cos x)[(0)(cos x) + (1)(sin x)] = 2(cos x)(sin x) = sin (2x)
The right angle triangle shows that you understand this intuition between translating sine to cosine geometrically. A simple graphical proof is too simplistic because you need to know this rock solid if you're going to be any good at calculus.
EDIT: Dunno if they still teach this, but it's the Indian chief "soh cah toa". Short for "sine = opposite/hypotenuse", "cosine = adjacent/hypotenuse", and "tangent = opposite/adjacent". Your right angle triangle should show that sine of angle x is the same as cosine from the (pi/2 - x) angle, meaning the other angle that is NOT the right angle.
|
sin(2x) = 2sin(x)cos(x)
2sin(x)cos(x)=2cos(x)*cos[pi/2-x]
sin(x)=cos(pi/2-x)
Now you show that x=pi/4 and that makes sin and cos equal.
On May 18 2015 09:43 coverpunch wrote:
I think the proof would be:
2 (cos x) [cos (pi/2-x)] = 2(cos x)[(cos pi/2)(cos x) + (sin pi/2)(sin x)] <=breaking it out using angle sum identity
2(cos x)[(cos pi/2)(cos x) + (sin pi/2)(sin x)] = 2(cos x)[(0)(cos x) + (1)(sin x)]
2(cos x)[(0)(cos x) + (1)(sin x)] = 2(cos x)(sin x) = sin (2x)
The right angle triangle shows that you understand this intuition between translating sine to cosine geometrically. A simple graphical proof is too simplistic because you need to know this rock solid if you're going to be any good at calculus.
This method, while it works, is overcomplicated. You just need to use a double angle formula (found here) to get the answer. From there it's relatively trivial. The sine/cosine curves intersect at pi/4 (45 degrees) and = rad(2)/2.
|
On May 18 2015 09:50 MtlGuitarist97 wrote:sin(2x) = 2sin(x)cos(x) 2sin(x)cos(x)=2cos(x)*cos[pi/2-x] sin(x)=cos(pi/2-x) Now you show that x=pi/4 and that makes sin and cos equal. Show nested quote +On May 18 2015 09:43 coverpunch wrote:
I think the proof would be:
2 (cos x) [cos (pi/2-x)] = 2(cos x)[(cos pi/2)(cos x) + (sin pi/2)(sin x)] <=breaking it out using angle sum identity
2(cos x)[(cos pi/2)(cos x) + (sin pi/2)(sin x)] = 2(cos x)[(0)(cos x) + (1)(sin x)]
2(cos x)[(0)(cos x) + (1)(sin x)] = 2(cos x)(sin x) = sin (2x)
The right angle triangle shows that you understand this intuition between translating sine to cosine geometrically. A simple graphical proof is too simplistic because you need to know this rock solid if you're going to be any good at calculus. This method, while it works, is overcomplicated. You just need to use a double angle formula (found here) to get the answer. From there it's relatively trivial. The sine/cosine curves intersect at pi/4 (45 degrees) and = rad(2)/2.
Sorry, but your solution is ridiculous. The identity works for any value of x, not just x = pi/4.
|
Alright, double angle formula was my plan it was just confusing because we were never actually taught that sin(x)=cos(pi/2-x)
This is probably something that is already evident or that I should be able to reason out easily but I am bad at this 
Mtlguitarist are you saying that both angles on my right triangle will be 45 degrees?
I guess they have to for the equation to work?
geeze im so bad at this lol. this is what I get for taking an online class
|
On May 18 2015 09:59 coverpunch wrote:Show nested quote +On May 18 2015 09:50 MtlGuitarist97 wrote:sin(2x) = 2sin(x)cos(x) 2sin(x)cos(x)=2cos(x)*cos[pi/2-x] sin(x)=cos(pi/2-x) Now you show that x=pi/4 and that makes sin and cos equal. On May 18 2015 09:43 coverpunch wrote:
I think the proof would be:
2 (cos x) [cos (pi/2-x)] = 2(cos x)[(cos pi/2)(cos x) + (sin pi/2)(sin x)] <=breaking it out using angle sum identity
2(cos x)[(cos pi/2)(cos x) + (sin pi/2)(sin x)] = 2(cos x)[(0)(cos x) + (1)(sin x)]
2(cos x)[(0)(cos x) + (1)(sin x)] = 2(cos x)(sin x) = sin (2x)
The right angle triangle shows that you understand this intuition between translating sine to cosine geometrically. A simple graphical proof is too simplistic because you need to know this rock solid if you're going to be any good at calculus. This method, while it works, is overcomplicated. You just need to use a double angle formula (found here) to get the answer. From there it's relatively trivial. The sine/cosine curves intersect at pi/4 (45 degrees) and = rad(2)/2. Sorry, but your solution is ridiculous. The identity works for any value of x, not just x = pi/4.
I wouldn't call it ridiculous just a mistake in what he's trying to do. forgot that he's trying to prove it in all cases and not just for x=pi/4 (which he seems to have pulled out for the sole purpose of making his formula work unless I'm missing something.)
so yeah he was solving for something when you need to do a full proof.
|
On May 18 2015 09:59 coverpunch wrote:Show nested quote +On May 18 2015 09:50 MtlGuitarist97 wrote:sin(2x) = 2sin(x)cos(x) 2sin(x)cos(x)=2cos(x)*cos[pi/2-x] sin(x)=cos(pi/2-x) Now you show that x=pi/4 and that makes sin and cos equal. On May 18 2015 09:43 coverpunch wrote:
I think the proof would be:
2 (cos x) [cos (pi/2-x)] = 2(cos x)[(cos pi/2)(cos x) + (sin pi/2)(sin x)] <=breaking it out using angle sum identity
2(cos x)[(cos pi/2)(cos x) + (sin pi/2)(sin x)] = 2(cos x)[(0)(cos x) + (1)(sin x)]
2(cos x)[(0)(cos x) + (1)(sin x)] = 2(cos x)(sin x) = sin (2x)
The right angle triangle shows that you understand this intuition between translating sine to cosine geometrically. A simple graphical proof is too simplistic because you need to know this rock solid if you're going to be any good at calculus. This method, while it works, is overcomplicated. You just need to use a double angle formula (found here) to get the answer. From there it's relatively trivial. The sine/cosine curves intersect at pi/4 (45 degrees) and = rad(2)/2. Sorry, but your solution is ridiculous. The identity works for any value of x, not just x = pi/4.
Oh, it doesn't just want two angles? Nvm then. Thought you just had to find a value of x where they were equal... Proofs are annoying. So glad I don't have to do trig proofs anymore.
On May 18 2015 10:01 travis wrote:
Mtlguitarist are you saying that both angles on my right triangle will be 45 degrees?
I guess they have to for the equation to work?
geeze im so bad at this lol. this is what I get for taking an online class
Nah I'm just bad at trig haha. I'd just take the advice of people who actually study math
|
OHHHH coverpunch I see what you are doing
but how do you think the teacher wants me to graphically represent this on the right triangle? just draw in x and pi/2-x as my angles? because that seems lame
|
On May 18 2015 10:03 travis wrote:
OHHHH coverpunch I see what you are doing
but how do you think the teacher wants me to graphically represent this on the right triangle? just draw in x and pi/2-x as my angles? because that seems lame
Yeah, and maybe label your triangle's sides a, b, and c (hypotenuse). So sin x = a/c and cos (pi/2-x) = a/c. Then you've proved it algebraically and you can see the geometric intuition in a picture.
EDIT: By the way, nothing is easily evident in math. It sweats the brain to think in this rigorous way.
|
Thank you coverpunch and everyone else. I am going to delete the blog now but I do appreciate the help.
oops nevermind they don't have a way for non forum staff to delete blogs. maybe one will later.
|
trig proofs are evil. i think it took me like 2-3 years of doing it before I actually figured it out somewhat (and now I've forgotten it because I don't do math anymore.)
gl on your final
|
Multiply two rotation matrices together
|
sin(2x)=IM(e^i2x)
e^i2x = e^ix * e^ix = (cos x + i sin x) * (cos x + i sin x) = cos^2 x - sin^2 x + i * 2 cos x sin x
IM(e^i2x) = 2 cos x sin x = 2 cos x cos (pi/2-x)
Wouldn't this be correct as well?
I'm not sure, it's been quite a while since i did this kind of math :$
|
Mute City2363 Posts
On May 18 2015 11:38 Yorbon wrote:
sin(2x)=IM(e^i2x)
e^i2x = e^ix * e^ix = (cos x + i sin x) * (cos x + i sin x) = cos^2 x - sin^2 x + i * 2 cos x sin x
IM(e^i2x) = 2 cos x sin x = 2 cos x cos (pi/2-x)
Wouldn't this be correct as well?
I'm not sure, it's been quite a while since i did this kind of math :$
This would be correct, but not too helpful in this case because this is basic trig, so travis probably hasn't got to imaginary numbers yet. Secondly, the key point in the question is showing that sin(x) = cos(pi/2 - x), which you've assumed in your answer.
Drawing the triangle is the way to go in pretty much every trig proof, and as soon as you get the right one then the answer is basically obvious from there.
On May 18 2015 09:01 travis wrote:
I could show any given 2 angles that add to 90 on the right triangle and label them x and (pi/2-x) but wtf would the point of that be?
This is a good start - the key point here is to think about what "2 angles adding up to 90" implies.
What's the 3rd going to be, and so how can you use your trig knowledge in this case?
Once you realise that the triangle to draw is right angled, then showing that sin(x) = cos(pi/2-x) should fall out immediately
|
Ah, apparently I misunderstood the question.
I don't think I've ever made a test where i couldn't even use sin(x) = cos(pi/2-x).
In that case this is indeed the case:
"the key point here is to think about what "2 angles adding up to 90" implies."
|
Random question but I saw you post a while back that you were living in Wichita, are you still living here? If so are you going to Wichita State?
|
On May 18 2015 12:52 GoShox wrote:
Random question but I saw you post a while back that you were living in Wichita, are you still living here? If so are you going to Wichita State?
no I moved from there about a year ago. I am going to university of maryland starting in july
|
|
|
|
|
|
EPT
