25 May 08
2 edits
Originally posted by kbaumenWell for a reality check, I did the calcs on my casio and finally convinced myself, any one of those worked out = 0.999 something and added together, it =2.999 or so. Don't see how that works out to be minus 1/2.
I had this as homework in my algebra course and it took a fair amount of time for me to find the solution. It's actually quite difficult to notice.
Prove the identity:
cos(2pi/7) + cos(4pi/7) + cos(6pi/7) = -1/2
Even if it was supposed to be inverted it would come out to 1/3 or so, a far cry from 1/2.
Are the units in degrees? or maybe radians?
-1/2 radian?
BTW, I am taking a course in calculus, on DVD's (4 of them) by 'the teaching company'. The prof is Michael Starbird Phd, and I really wished I had a teacher that good when I was in college, this guy is incredible, with new results from recent mathemeticians and excellent visuals and step by step explainations of the concepts behind like derivitives and integrals and such, overall, a GREAT course, highly recommended. If you take that course before you get to college you will be well on your way to an A in calculus!
Back to the cosign: 2Pi is 6.28 and /7= 0.897 and cos (0.8970) = 0.999 any way you look at it and making it 2PI or 4PI or 6PI doesn't change it much, only in the 4th and 5th digit, so can't see adding those three values and coming up with anything like 1/2 or -1/2.
25 May 08
2 edits
Originally posted by sonhouseEverywhere with pi it's radians - 2pi/7 radians, 4pi/7 radians, etc. Isn't that basic notation in trigonometry? To simply write a number without a superscript '0' to point out that those are radians?
Well for a reality check, I did the calcs on my casio and finally convinced myself, any one of those worked out = 0.999 something and added together, it =2.999 or so. Don't see how that works out to be minus 1/2.
Even if it was supposed to be inverted it would come out to 1/3 or so, a far cry from 1/2.
Are the units in degrees? or maybe radians?
-1/2 rad igit, so can't see adding those three values and coming up with anything like 1/2 or -1/2.
-1/2 is a number.
It's a normal problem. I don't think that 120 pupils (ok, 60, some couldn't prove it) and 2 teachers could altogether make a single mistake and prove the unprovable.
EDIT: Btw, I checked it with a calculator (basic Gnome calculator) and it really came out -1/2. So you probably calculated in degrees.
25 May 08
2 edits
Originally posted by sonhouseYou should try it again, but this time using RADIAN measure (which is what is implied when giving the cos of some factor of PI. (And it works out perfectly using RADIAN measure.)
Back to the cosign: 2Pi is 6.28 and /7= 0.897 and cos (0.8970) = 0.999 any way you look at it and making it 2PI or 4PI or 6PI doesn't change it much, only in the 4th and 5th digit, so can't see adding those three values and coming up with anything like 1/2 or -1/2.
EDIT: Also, this does not seem like an IDENTITY to me. An identity is something like cos^2(x) + sin^2(x) = 1, where x can be any value. This is a simple, 'type-it-into-your-calculator-and-check-to-see-if-it-is-in-radian-measure' type question, that we math teachers love to give.
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25 May 08
Originally posted by kbaumenI believe you that it is an Identity, but i havent worked with pi/7 in any that I have ever done ( witch isnt many) your probably not going to want decimal equivalents of the radians, since it is such a clean target. i cant think of a way to use cos(t1+/- t2) to find the cosine of a 7th radian......
Everywhere with pi it's radians - 2pi/7 radians, 4pi/7 radians, etc. Isn't that basic notation in trigonometry? To simply write a number without a superscript '0' to point out that those are radians?
-1/2 is a number.
It's a normal problem. I don't think that 120 pupils (ok, 60, some couldn't prove it) and 2 teachers could altogether make a single mist ...[text shortened]... sic Gnome calculator) and it really came out -1/2. So you probably calculated in degrees.
26 May 08
Originally posted by GastelYep, right of course. I thought that's what my problem was, but I was having trouble getiing my little casio to go into radian mode, finally figured that one out, and it came out at -1/2 (actually -0.5)
You should try it again, but this time using RADIAN measure (which is what is implied when giving the cos of some factor of PI. (And it works out perfectly using RADIAN measure.)
EDIT: Also, this does not seem like an IDENTITY to me. An identity is something like cos^2(x) + sin^2(x) = 1, where x can be any value. This is a simple, 'type-it-into-your-ca ...[text shortened]... nd-check-to-see-if-it-is-in-radian-measure' type question, that we math teachers love to give.
The thing that puzzles me is with one radian = 57 odd degrees, 180/PI, why can't I convert the calc using degrees into radians at the end? The numerical answer at around 3, first off is not negative. Oh, I guess the - part comes from the angle being to the left of high noon (the zero degree hack). Is that right? So the -sign doesn't even mean the same thing as an algebaic - sign. I think, correct me if I am in left field here. So anyway, 0.5 radians is about 28 degrees. How do you convert the answer I got doing the calcs with degrees, which came out around 3? Three degrees is more like 0.05 radians not 0.5.
I must be bolluxing the units then. Help!
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26 May 08
2 edits
Originally posted by sonhousemultiply degrees by Pi/180 to convert from degrees to radians, and vice versa for radians to degrees
Yep, right of course. I thought that's what my problem was, but I was having trouble getiing my little casio to go into radian mode, finally figured that one out, and it came out at -1/2 (actually -0.5)
The thing that puzzles me is with one radian = 57 odd degrees, 180/PI, why can't I convert the calc using degrees into radians at the end? The numerical a Three degrees is more like 0.05 radians not 0.5.
I must be bolluxing the units then. Help!
You have to think about the Cartesian plane......if the cosine is negative the angle resides in the second or third quadrant then the angle is obtuse either way
if i am understanding the question
I think your calcs are incorrect a cosine of -1/2 belong to an angle of 120 degrees, or 2pi/3 radians and 240 degrees or 4pi/3 radians in the range of ( 0 , 2Pi )
so the whole left side should be able to be manipulated into somthing of the sort using known identities, but it is tough
26 May 08
Originally posted by kbaumenI'm aware of a variation of this problem that was included in the 1963 math olympics. So it is indeed a tough problem to solve:
I had this as homework in my algebra course and it took a fair amount of time for me to find the solution. It's actually quite difficult to notice.
Prove the identity:
cos(2pi/7) + cos(4pi/7) + cos(6pi/7) = -1/2
Prove:
cos pi/7 - cos 2pi/7 + cos 3pi/7 = 1/2
26 May 08
1 edit
Originally posted by GastelThis is an identity. A trigonometric identity is an equality that involves trigonometric functions. What you have provided is is a basic relationship used to prove other identities. Similar to tgx = sinx/cosx.
EDIT: Also, this does not seem like an IDENTITY to me. An identity is something like cos^2(x) + sin^2(x) = 1, where x can be any value. This is a simple, 'type-it-into-your-calculator-and-check-to-see-if-it-is-in-radian-measure' type question, that we math teachers love to give.
Now, can anyone prove it? (The one I provided - cos(2pi/7) + ...)
26 May 08
1 edit
The problem is to prove LHS=RHS, which can be done..I'll reply with the answer soon..As far as i can remember u have to prove another known identity first and get a 4th degree equation in cos(x) where x = 2n(pi)/7 and n is an integer..so the roots of that equation wil be cos(0),cos(2pi/7),cos(4pi/7) and cos(6pi/7).. then u can directly get the desired equation by getting the value for the summation of the roots..i.e. like in a quadratic equation(ax2+bx+c=0) where alpha+beta = -(b/a)..hope this helps.. 😀
26 May 08
Originally posted by blacknight1985ok here it is..the other identity is when x=2n(pi)/7, cos(3x)=cos(4x)..(u can prove it urself easily) then u expand both sides to get the desired 4th degree equation i've mentioned earlier..then the rest is the same..
The problem is to prove LHS=RHS, which can be done..I'll reply with the answer soon..As far as i can remember u have to prove another known identity first and get a 4th degree equation in cos(x) where x = 2n(pi)/7 and n is an integer..so the roots of that equation wil be cos(0),cos(2pi/7),cos(4pi/7) and cos(6pi/7).. then u can directly get the desired ...[text shortened]... ..i.e. like in a quadratic equation(ax2+bx+c=0) where alpha+beta = -(b/a)..hope this helps.. 😀
26 May 08
Originally posted by kbaumenWhat you have here is an equality which needs proving NOT an identity which as previously discussed will involve a variable.
This is an identity. A trigonometric identity is an equality that involves trigonometric functions. What you have provided is is a basic relationship used to prove other identities. Similar to tgx = sinx/cosx.
Now, can anyone prove it? (The one I provided - cos(2pi/7) + ...)
eg x^2 - y^2 = (x-y)(x+y) is an identity
That of course doesnt make it any easier! 😞
26 May 08
Having tried various different approaches - and proving various identities that weren't asked for (I'm seriously out of practice with these things) - I've got a method that's so simple I suspect there's something wrong with it!
[Will write 2pi/7 as a, to simplify things]
It relies on e^(ia) being a root of x^7 = 1.
But x^7 = (x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)
=> e^(6ia) + e^(5ia) + e^(4ia) + e^(3ia) + e^(2ia) + e^(ia) + 1 = 0
Taking the real part of this:
cos(6a) + cos(5a) + cos(4a) + cos(3a) + cos(2a) + cos(a) + 1 = 0
But note: cos(2pi - x) = cos(x)
so cos(6a) = cos(12pi/7) = cos(2pi/7) = cos(a)
and cos(5a) = cos(2a)
and cos(4a) = cos(3a)
=> 2[cos(a) + cos(2a) + cos(3a)] + 1 = 0
=> cos(2pi/7) + cos(4pi/7) + cos(6pi/7) = -1/2
Ta-da!