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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
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okay so then I guess the problem must really be that simple
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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.
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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.
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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.
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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.
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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
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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.
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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
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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
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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.
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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.
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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
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Multiply two rotation matrices together
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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 :$
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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
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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."
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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?
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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
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Hello, travis.
I've written a solution to your problem. Unfortunately, I can't post the .pdf file here, so I had to export it to two lesser quality image files. Here they are.
![[image loading]](http://i.imgur.com/lWIDU4C.png)
![[image loading]](http://i.imgur.com/iPuVPN0.png)
In addition, here are some other trigonometric identities I think you'll find useful in the future.
![[image loading]](http://i.imgur.com/UhKojck.png)
Good luck on your final.
Sincerely,
Shalashaska_123
EDIT: Fixed some mistakes. 
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Nice write-up, but sin(x) = cos(pi/2-x) is pretty much evident from the definitions of sinus and cosinus, so I don't know what's to discuss.
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It sticks better and represents a grasp of the intuition if you're able to derive it algebraically.
In US colleges, stuff like this also tends to be a way of weeding out untalented or unmotivated students. If you can't derive this using the other identities and/or you don't want to try, then you don't have the ability or work ethic to succeed in a mathematical career. Both because the material gets much harder when you hit analysis, and it's masochistic to try if you don't know how these fundamental things are related to each other.
I am curious about something. American math books tend to have explanations with "proof is left as an exercise to the reader". I'll assume the big names for standard textbooks are the same in all English speaking countries. Do they do this with math books in other languages? I would tentatively guess yes, having studied books translated into English by German or Russian authors and they do it too, but I'd just like to know from a person who actually uses books in other languages.
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Well I can say that it happens in Danish math books as well although only sparsely - my experience is really mostly with math through high-school (and a lot of statistics at university, but that was in English). At our universities we generally use English books, so it's the same as in the UK/US.
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Mute City2363 Posts
On May 18 2015 20:06 Maenander wrote:
Nice write-up, but sin(x) = cos(pi/2-x) is pretty much evident from the definitions of sinus and cosinus, so I don't know what's to discuss.
You can say that for everything in Maths. Everything is trivial to someone who understands it thoroughly. The important part here is teaching bulletproof logic.
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stealing this image:
sin x = b /c
cos ( .5 pi - x ) = b /c
:.sin x = cos ( .5 pi - x )
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May I ask how old you are and the level at which this is taught?
On May 18 2015 20:06 Maenander wrote:
Nice write-up, but sin(x) = cos(pi/2-x) is pretty much evident from the definitions of sinus and cosinus, so I don't know what's to discuss.
Exactly
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By the way, I forgot to mention in my article that even though the quantity sin 2x = 2ab/c^2 involves two distances multiplied together in the numerator and denominator, the ratio itself is dimensionless as is the case for any trigonometric function. When you verify different identities using the right triangle, keep in mind the fact that there should never be more distances multiplied together in the numerator than in the denominator or vice-versa, or else you have done something wrong.
EDIT: Also, if anyone would like to have more practice with these kinds of problems, try proving the identities in the right column of the handout I posted with the identities in the left column.
On May 18 2015 21:38 thecrazymunchkin wrote:Show nested quote +On May 18 2015 20:06 Maenander wrote:
Nice write-up, but sin(x) = cos(pi/2-x) is pretty much evident from the definitions of sinus and cosinus, so I don't know what's to discuss. You can say that for everything in Maths. Everything is trivial to someone who understands it thoroughly. The important part here is teaching bulletproof logic.
No, Maenander is right. As Cambium eloquently wrote,
On May 18 2015 22:24 Cambium wrote:
sin x = b /c
cos ( .5 pi - x ) = b /c
:.sin x = cos ( .5 pi - x )
it is quite simple to show that sin x = cos ( .5 pi - x ) from the definitions of the functions. I wouldn't go so far as to say there's nothing to discuss, though. Clearly there are some people, such as travis, that are confused and need some explanation. That's where I think you are right.
On May 18 2015 22:24 Cambium wrote:
stealing this image:
Oh, are you in the same class as travis? That's nice I was able to kill two birds with one stone. Show my second page some love, too!
On May 18 2015 20:40 coverpunch wrote:
It sticks better and represents a grasp of the intuition if you're able to derive it algebraically.
Yes, I agree. On exams especially, where you need to demonstrate your knowledge to the professor, the geometric reasoning isn't nearly as impressive. One should ideally be able to think about these problems both algebraically and geometrically.
On May 18 2015 20:40 coverpunch wrote:
In US colleges, stuff like this also tends to be a way of weeding out untalented or unmotivated students. If you can't derive this using the other identities and/or you don't want to try, then you don't have the ability or work ethic to succeed in a mathematical career. Both because the material gets much harder when you hit analysis, and it's masochistic to try if you don't know how these fundamental things are related to each other.
Analysis is definitely the class that distinguishes the thinkers from the calculators in the mathematics curriculum. Unfortunately, what one is taught in grade school in the U.S. doesn't prepare students for this class--the emphasis is on computing. For example, you're given the Pythagorean Theorem or Law of Cosines and asked to compute sides of triangles. You're told the results of theorems and asked to compute a derivative or solve differential equations with integration or series solutions without any questions about the theorems themselves. Most if not all of the questions in grade school involve lots of tedious calculations with little to no creative thinking on the part of the student. Mathematics is really concerned with "why" as opposed to "how" things work, and it's a shame that students (at least mathematics students in the U.S.) don't get a good idea of this early on. Anyway, pardon my little off-topic rant. I just wanted to say what was on my mind.
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On May 19 2015 03:31 Grovbolle wrote:May I ask how old you are and the level at which this is taught? Show nested quote +On May 18 2015 20:06 Maenander wrote:
Nice write-up, but sin(x) = cos(pi/2-x) is pretty much evident from the definitions of sinus and cosinus, so I don't know what's to discuss. Exactly
I am 30 years old, I am being taught this at community college. I had to start with "intermediate algebra", which was a pre-req class. Then a pre-calculus class that was basically algebra 2, then this pre-calculus class which is basically trig.
This class was an online class, which is really just code for self-taught, so my understanding is a lot weaker than the other 2 classes... because I did a pretty unmotivated job of teaching myself.
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Mute City2363 Posts
On May 19 2015 06:30 Shalashaska_123 wrote:Show nested quote +On May 18 2015 21:38 thecrazymunchkin wrote:On May 18 2015 20:06 Maenander wrote:
Nice write-up, but sin(x) = cos(pi/2-x) is pretty much evident from the definitions of sinus and cosinus, so I don't know what's to discuss. You can say that for everything in Maths. Everything is trivial to someone who understands it thoroughly. The important part here is teaching bulletproof logic. No, Maenander is right. As Cambium eloquently wrote, Show nested quote +On May 18 2015 22:24 Cambium wrote:
sin x = b /c
cos ( .5 pi - x ) = b /c
:.sin x = cos ( .5 pi - x ) it is quite simple to show that sin x = cos ( .5 pi - x ) from the definitions of the functions. I wouldn't go so far as to say there's nothing to discuss, though. Clearly there are some people, such as travis, that are confused and need some explanation. That's where I think you are right.
You're right, I didn't really express myself correctly. It would be more accurate to say that nothing is trivial to someone who doesn't understand it, which is why I think that dismissing it as 'evident' in this case was a little counterproductive. In any case, hopefully we've got there in the end.
Do you teach in the States then? I feel we agree on quite a lot of the flaws of the maths teaching system; nice (or not nice, really) to know that it's not just a British problem!
On May 19 2015 10:27 travis wrote:
I am 30 years old, I am being taught this at community college. I had to start with "intermediate algebra", which was a pre-req class. Then a pre-calculus class that was basically algebra 2, then this pre-calculus class which is basically trig.
This class was an online class, which is really just code for self-taught, so my understanding is a lot weaker than the other 2 classes... because I did a pretty unmotivated job of teaching myself.
If you need some stuff explaining feel free to give me a shout anytime
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doubleupgradeobbies!
Australia1187 Posts
This problem becomes trivially easy, even in an algebraic sense if you use the Euler's representation of sine and cosine. Sadly, I'm not sure if your professor would accept it, but it's by far the most elegant non-graphical proof I can think of off the top of my head.
2*cos(x)cos(pi/2-x) = 2*((e^ix +e^-ix)/2)) * ((e^(pi*i/2 - ix) + e^-(pi*i/2-ix)/2))
= (e^ix +e^-ix)*((ie^-ix - ie^ix)/2)
= (e^i2x - e^-i2x)/2i
= sin(2x)
I know alot of professors don't like to accept answers outside the scope of the course, but seriously, how easy is that?
edit: oops was originally proving the wrong thing
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On May 18 2015 09:01 travis wrote:I am reviewing for my trig final (THIS ISNT AN ASSIGNMENT, IT IS REVIEW). Two of the questions in the review aren't really something that was directly covered in the class, and I am having trouble understanding it and also trouble researching it myself. The problem says: Show nested quote +
prove the identity
sin(2x) = 2 (cos x) [cos (pi/2-x)]
then use a right triangle to show angles x and (pi/2 - x)
I don't really understand what my professor is asking me to do. I mean, I know sin(2x) = 2 cos(x)(sin x) so then i guess sin(x) = cos(pi/2-x) ? But I mean I don't really know how to prove that, or what my professor wants me to *show* on the right triangle. 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? P.S. I'll delete this blog after some time or after I get help because I know how strict TL can be about shitty blogs
This is how I understood the problem. You have to show geometrically, using right triangles that the above formula is correct.
![[image loading]](http://i.imgur.com/3AelcUc.jpg)
This geometrical "proof" (can't prove anything with drawings) uses:
Center angle is twice as big as angle from a point on the circle
Triangles with points on a circle and one side as a diameter are right triangles
In all triangles, sin(Â)/BC=sin(B^)/AC=sin(C^)/AB
Sum of angles in a triangle is pi
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Another question is whether you need to prove the double-angle formula, or any other sum-difference formulas that many people used in their proofs - or are they expected that you memorize and use them? In which case a proof like what Geiko did is necessary, although he did use the law of sines. Also he didn't completely finish at the end; just convert sin(pi/2-a) = cos(pi/2 - pi/2 +a) = cos(a) and sin(a) = cos(pi/2 - a) completing the proof.
For people using Euler's formula that might need to be proven using a power series as well, otherwise there's no good reason to accept the formula (it is the foundation for one of the most famous equalities in math after all, it probably deserves an explanation). Given that they probably are far far away from power series I don't think its reasonable to use such proofs at that level.
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sin x = cos ( .5 pi - x ) from the definitions of the functions being obvious clearly depends of the definitions. Have fun doing that from the power series. Hard or easy rarely is clear cut.
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For people using Euler's formula that might need to be proven using a power series as well, otherwise there's no good reason to accept the formula (it is the foundation for one of the most famous equalities in math after all, it probably deserves an explanation). Given that they probably are far far away from power series I don't think its reasonable to use such proofs at that level.
Yes, it's a good assumption that you should never use something outside of the requirements to take the course and the course itself in math. The instructor wouldn't know whether you have covered the material or you found a formula and blindly applied it. And using things that you don't understand kind of defeats the point of the courses, and in a way, math itself.
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On May 20 2015 05:34 radscorpion9 wrote:
Another question is whether you need to prove the double-angle formula, or any other sum-difference formulas that many people used in their proofs - or are they expected that you memorize and use them? In which case a proof like what Geiko did is necessary, although he did use the law of sines. Also he didn't completely finish at the end; just convert sin(pi/2-a) = cos(pi/2 - pi/2 +a) = cos(a) and sin(a) = cos(pi/2 - a) completing the proof.
For people using Euler's formula that might need to be proven using a power series as well, otherwise there's no good reason to accept the formula (it is the foundation for one of the most famous equalities in math after all, it probably deserves an explanation). Given that they probably are far far away from power series I don't think its reasonable to use such proofs at that level.
From my experience in American math classes, at the early levels they'll give you a sheet of identities like the one Shalashaska provided and you're expected to "plug and chug" as they say. From the OP's subsequent posts, it appears cos(pi/2-x) = sin(x) was not one of the identities he was given so he was expected to prove it algebraically and show it geometrically.
It's only after you demonstrate a firm grasp of the basics and move on to basic analysis courses or other courses developing proofs that you'll go back and do proofs for the identities. I did my Master's in mathematical statistics and one memorable problem was going back and doing a proof for Hardy-Weinberg equilibrium (that in a population with two alleles and random mating, every generation will have genotypes in ratio AA:2Aa:aa). In high school biology, you learn to use Punnett squares with Mendel's laws and see the result with counting. It's much trickier to derive the result algebraically and generalize it to any number of subsequent generations. But at that level, it was a useful exercise to teach statisticians how to approach problems in other subjects from a mathematical and theoretical perspective.
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On May 20 2015 05:34 radscorpion9 wrote:
Another question is whether you need to prove the double-angle formula, or any other sum-difference formulas that many people used in their proofs - or are they expected that you memorize and use them? In which case a proof like what Geiko did is necessary, although he did use the law of sines. Also he didn't completely finish at the end; just convert sin(pi/2-a) = cos(pi/2 - pi/2 +a) = cos(a) and sin(a) = cos(pi/2 - a) completing the proof.
For people using Euler's formula that might need to be proven using a power series as well, otherwise there's no good reason to accept the formula (it is the foundation for one of the most famous equalities in math after all, it probably deserves an explanation). Given that they probably are far far away from power series I don't think its reasonable to use such proofs at that level.
radscorpion9, you don't have to re-invent the wheel to write an acceptable proof. The sine double angle formula and the cosine difference formula that I used are presented as theorems and proven in any precalculus text. These are elementary, and you don't have to prove them every time you use them. There are better things to do with your time; in addition, it takes the focus away from the problem at hand. If you happen to forget them and don't have a reference like the handout i provided, then yes, you'll have to re-derive them. The same goes for the Law of Sines. I'll briefly go over the proofs in case you are curious..
Sine Double Angle Formula from Sine Sum Formula
sin 2x = sin (x + x) = sin x cos x + cos x sin x = 2sin x cos x
Sine Sum formula from Cosine Difference Formula
sin (a + b) = cos ( pi/2 - a - b ) = cos( pi/2 - a )cos( b ) + sin( pi/2 - a )sin(b) = sin a cos b + cos a sin b
Outline of Cosine Difference Formula Proof
Consider two points P_1 and P_2 on a unit circle that have angles a and b, respectively, with respect to the x-axis. Rotate the circle clockwise by an angle b so that the points that were at P_1 and P_2 are now at the new points P_3 (with angle a - b with respect to x-axis) and P_4 (with angle 0 with respect to x-axis), respectively. The distance between P_1 and P_2 is equal to that between P_3 and P_4. Simplify this equation with tricks from algebra, and the formula, cos (a - b) = cos a cos b + sin a sin b, pops out.
Your second paragraph is nonsense. Of course there are good reasons to accept Euler's formula. If you're faced with the task of simplifying a nasty trigonometric expression that even that handout can't help you with, it's often far more convenient to convert everything to exponentials because of their handsome algebraic properties. Once you're done working with exponentials, switch back to trigonometric functions as doubleupgradeobbies! did in his or her proof.
On May 19 2015 15:49 doubleupgradeobbies! wrote:
2*cos(x)cos(pi/2-x) = 2*((e^ix +e^-ix)/2)) * ((e^(pi*i/2 - ix) + e^-(pi*i/2-ix)/2))
= (e^ix +e^-ix)*((ie^-ix - ie^ix)/2)
= (e^i2x - e^-i2x)/2i
= sin(2x)
As I said in my previous post, the education system emphasizes computation rather than proofs. Students don't learn mathematical induction and prove De Moivre's formula before they start using it to find complex roots. Similarly, students don't prove Euler's formula with Taylor series from calculus before they start using it to manipulate trigonometric expressions.
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Mute City2363 Posts
To quote the OP:
On May 19 2015 10:27 travis wrote:
I am 30 years old, I am being taught this at community college. I had to start with "intermediate algebra", which was a pre-req class. Then a pre-calculus class that was basically algebra 2, then this pre-calculus class which is basically trig.
I highly doubt that he's got to complex numbers yet. There's no point in not keeping it simple for now
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On May 20 2015 16:04 thecrazymunchkin wrote:To quote the OP: Show nested quote +On May 19 2015 10:27 travis wrote:
I am 30 years old, I am being taught this at community college. I had to start with "intermediate algebra", which was a pre-req class. Then a pre-calculus class that was basically algebra 2, then this pre-calculus class which is basically trig. I highly doubt that he's got to complex numbers yet. There's no point in not keeping it simple for now
Complex numbers are covered in Algebra 2 here in the U.S., but you're right. It's always best to keep it as simple as possible. I was just mentioning the exponential approach above in case a much harder problem than the one dealt with here was encountered. With that being said,
On May 20 2015 06:53 corumjhaelen wrote:
sin x = cos ( .5 pi - x ) from the definitions of the functions being obvious clearly depends of the definitions. Have fun doing that from the power series. Hard or easy rarely is clear cut.
use the SOH-CAH-TOA definitions. Why would you make life difficult for yourself using series representations of the functions? Taylor series shouldn't even come to mind when dealing with such a simple problem in trigonometry. They wouldn't even help a trigonometry student either.
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EPT
