How to prove cos36 - cos72 = .5
ykm
Please help!
Jon
On Tue, 02 Jul 1996 00:49:48 GMT, k...@hk.super.net (ykm) wrote:
>Please help!
>
Sure:
cos(36)=sqrt(5)/4 +1/4
cos(72)=sqrt(5)/4 - 1/4
so:
cos(36)-cos(72) = 1/2
isn't math neat!
Life is grand, don't mess it up..
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Peter L. Montgomery
In article <4r9v3q$8...@tst.hk.super.net> k...@hk.super.net (ykm) writes:
>Please help!
Please repeat your question in the body of your posting.
Some newsreaders truncate the subject line, although that was
not a problem this time.
With all angles in degrees,
cos(36) - cos(72)
= 2 * sin(18) * sin(54)
= 0.5 * sin(2*18) * sin(2*54) / (cos(18) * cos(54))
= 0.5 * sin(36) * sin(108) / (sin(108) * sin(36))
= 0.5.
Or, if x = cos(36) - cos(72), then
2x^2 = 2*cos(36)^2 - 4 * cos(36) * cos(72) + 2*cos(72)^2
= (cos(72) + 1) - 2*(cos(36) + cos(108)) + (cos(144) + 1)
= (cos(72) + 1) - 2*(cos(36) - cos(72)) + (-cos(36) + 1)
= 2 + 3*cos(72) - 3*cos(36) = 2 - 3x.
This quadratic has two roots, one of which is easily eliminated.
--
Peter L. Montgomery pmon...@cwi.nl San Rafael, California
My sex is male. I am ineligible for an abortion.
Kurt Foster
ykm (k...@hk.super.net) wrote:
: Please help!
:
Let u = 2*cos(2*pi/5). Consider the isosceles triangle with angles
2*pi/5, 2*pi/5 and pi/5, the two equal sides having length 1. The base
then has length 2*sin(pi/10) = 2*cos(2*pi/5) = u. The angle bisector of
one of the two equal angles then divides the opposite side into parts of
lengths u and u^2, whence
1) u^2 + u = 1
From u = 2*cos(2*pi/5) we obtain from the double-angle formula
2) u^2 - 2 = 2*cos(4*pi/5). Since cos(pi - x) = - cos(x) we have
3) u^2 - 2 = -cos(pi/5).
Multiplying both sides of (1) by -1, then adding 2 to both sides gives
4) -(u^2 - 2) - u = 1. Substituting for u and u^2 - 2 gives
5) 2*cos(pi/5) - 2*cos(2*pi/5) = 1. Dividing by 2 gives the required
result.
Travis Kidd
Well, cos 36 = sqrt((3+sqrt(5))/8)
and cos 72 = sqrt((3-sqrt(5))/8).
So cos36 - cos72 = sqrt((3+sqrt(5))/8 + (3-sqrt(5))/8
- 2*sqrt((3+sqrt(5))(3-sqrt(5))/64))
= sqrt(6/8 - 2*sqrt(9-5)/8))
= sqrt(3/4 - 2*sqrt(4)/8)
= sqrt(3/4 - 2/4)
= sqrt(1/4)
= +/- .5
Finally, since cos36 > cos72, this must equal .5
QED.
Better challenges for ya:
Prove: sin x + sin (60-x) == sin (60+x)
and
(sin x)(sin(60-x))(sin(60+x)) == sin(3x).
-Travis
Michael Clark
In article <4r9v3q$8...@tst.hk.super.net>, ykm
<mailto:k...@hk.super.net> wrote:
>
> Please help!
>
Use the formula: CosA - CosB = -2(sin((A + B)/2))(sin((A - B)/2))
--
Michael Clark Maths Teacher, Acorn RiscPC and Pocketbook User.
There are lies, damned lies and Computer Industry "Standards".
Richard Mathar
In article <4r9v3q$8...@tst.hk.super.net>, k...@hk.super.net (ykm) writes:
...
|> Please help!
|>
If we set 36 degrees = p = pi/5 rad the equation to prove is
cos(p)-cos(2p)=1/2
Using cos(2p) = 2*cos^2(p)-1 this is equivalent to
cos(p)-2*cos^2(p)+1=1/2
or with P:= cos(p)
P^2-P/2-1/4=0
with the solutions P=1/4 +- sqrt(5)/4
It is left to prove that cos(pi/5) is one of these 2 numbers.
We wish to compute cos(p)=cos(pi/5)
and know sin(pi)=0=sin(5*p)
and know cos(5*p)=sqrt[1-sin^2(5*p)]
Therefore with p=pi/5 (pi=3.1415926...)
and sin(5*p)=sin(p)*[16*cos^4(p)-12*cos^2(p)+1]=0
and omitting the solutions with sin(p)=0 that are obviously not searched for
we have
cos^4(p)-(3/4)cos^2(p)+1/16=0
which is a quadratic equation for cos^2(p):
cos^2(p)=3/8+-sqrt( 5/64)=3/8+-sqrt(5)/8
It is left to show that +-sqrt(3/8+-sqrt(5)/8) is P as given above.
This can be qualitatively done by setting both numbers equal and building the
square...
Shimpei Yamashita
Travis Kidd <tk...@hubcap.clemson.edu> writes:
>
>k...@hk.super.net (ykm) writes:
>>Please help!
>
>Well, cos 36 = sqrt((3+sqrt(5))/8)
> and cos 72 = sqrt((3-sqrt(5))/8).
Oh, and where did you get *those* numbers, my friend?
For complete credit, you need to derive the above result as
well. (There is a neat geometrical proof, but I'd like to see what
other clever proofs people can come up with....)
>So cos36 - cos72 = sqrt((3+sqrt(5))/8 + (3-sqrt(5))/8
> - 2*sqrt((3+sqrt(5))(3-sqrt(5))/64))
>
> = sqrt(6/8 - 2*sqrt(9-5)/8))
>
> = sqrt(3/4 - 2*sqrt(4)/8)
>
> = sqrt(3/4 - 2/4)
>
> = sqrt(1/4)
>
> = +/- .5
>
>Finally, since cos36 > cos72, this must equal .5
>
> QED.
>
>Better challenges for ya:
>
>Prove: sin x + sin (60-x) == sin (60+x)
sin(60-x)=sin(60)cos(x)-cos(60)sin(x)
=sqrt(3)/2 cos(x) - 1/2 sin(x)
sin(x)+sin(60-x)=sqrt(3)/2 cos(x) + 1/2 sin(x)
=sin(60) cos(x) + cos(60) sin(x)
=sin(60+x)
> and
>
> (sin x)(sin(60-x))(sin(60+x)) == sin(3x).
Obviously false. (Try x=90. Left side is -1/4, right side equals -1. Oops!)
sin(x)(sin(60-x)sin(60+x))
=sin(x)(sin(60)cos(x)-cos(60)sin(x))(sin(60)cos(x)+cos(60)sin(x))
=sin(x)(sin^2(60)cos^2(x)-cos^2(60)sin^2(x))
=sin(x)(3/4 cos^2(x)- 1/4 sin^2(x))
=sin(x)(3/4 - sin^2(x))
=3/4 sin(x) - sin^3(x) != sin(3x)
--
Shimpei Yamashita <http://www-leland.stanford.edu/%7Eshimpei/>
Michael Press
It is straight forward to show (see below)
cos((k+1) * x) = 2 * cos(x) * cos (k * x) - cos ((k-1)* x).
Therefore
cos (2x) = 2 * (cos (x))^2 - 1
cos (3x) = 4 * (cos (x))^3 - 3 * cos (x)
cos (4x) = 8 * (cos (x))^4 - 8 * (cos (x))^2 + 1
cos (5x) = 16 * (cos (x))^5 - 20 * (cos (x))^3 + 5 * cos (x)
Since cos (5 * 18) = cos (90) = 0,
g := cos (18 degree) is a root of 16X^4 - 20 X^2 + 5. Therefore
2 * (cos (36) - cos (72)) = 2 * (2g^2 -1 - (8g^4 - 8g^2 + 1))
= -16g^4 + 20g^2 - 4
= 5 - 4.
Proof of the recurrence formula for cos ((k+1)* x):
Let M denote the matirx | cos x -sin x |
| sin x cos x |
and let T_k denote the vector | cos (k * x) |
| sin (k * x) |.
We have M^2 = 2 * cos (x) * M - I, and
the addition formulae for sin and cos give
T_k = M * T_k-1 so that
T_k+1 = M^2 * T_k-1
= 2 * cos (x) * M * T_k-1 - T_k-1
= 2 * cos (x) * T_k - T_k-1.
and the result we seek is the equation obeyed by the first
components of the vectors in this formula.
--
Michael Press
pre...@apple.com
Herman Serras
Michael Press wrote:
>
> It is straight forward to show (see below)
> cos((k+1) * x) = 2 * cos(x) * cos (k * x) - cos ((k-1)* x).
> Therefore
> cos (2x) = 2 * (cos (x))^2 - 1
> cos (3x) = 4 * (cos (x))^3 - 3 * cos (x)
> cos (4x) = 8 * (cos (x))^4 - 8 * (cos (x))^2 + 1
> cos (5x) = 16 * (cos (x))^5 - 20 * (cos (x))^3 + 5 * cos (x)
>
>
> Michael Press
> pre...@apple.com
Why not use a little geometry? Inscribing a regular decagon in a circle
gives a midpoint angle of 36 degrees. It's well known that the side of
this decagon is the largest part of the two parts in which the radius is
divided by the golden section. So, if the radius is 1 than the side of
the decagon is (sqrt(5)-1)/2. So it's easy to obtain
sin 18(degrees)=(sqrt(5)-1)/4.
Expressing cos 36 - cos 72 in sin 18 (elementary trigonometry) gives the
answer.
Herman Serras
Herman...@rug.ac.be
Hauke Reddmann
Let sin(18)=cos(72)=a,sin(36)=b,cos(36)=c,sin(72)=cos(18)=d.
By the simplest angle-doubling formula:
b=2ad c=d^2-a^2 d=2bc a=c^2-b^2 and of course a^2+d^2=b^2+c^2=1.
So a-c=(2c^2-1)-(1-2a^2) and bd=2ad*2bc, so 1=4ac.
Subtract to get a-c+1=2*(a-c)^2. This has solutions a-c=1,-1/2.
As a-c is negative,it is -1/2. Q.E.D.
Now for something completely different:
cos(36)*cos(72)=1/4
--
Hauke Reddmann <:-EX8
fc3...@math.uni-hamburg.de PRIVATE EMAIL
fc3...@rzaixsrv1.rrz.uni-hamburg.de BACKUP
redd...@chemie.uni-hamburg.de SCIENCE ONLY
David Petry
Here's a way.
let a = exp(i pi / 5) so a^5 = -1
then a* (1/a + 1)/(a + 1) = 1 = a* (1 - a^4)/(a+1)
= a - a^2 + a^3 - a^4 = a - a^2 - 1/a^2 + 1/a
= 2*cos(pi/5) - 2*cos(2 pi/5)
Herman Serras
Some days ago I suggested to use a little geometry. I think following
the same way I have a nicer proof!
Starting with the same regular decagon inscibed in the unit circle and
with the knowledge that the side of this decagon is the largest part of
the two parts in which the radius is divided by the golden section we
get:
1/(2 sin(18)) = 2 sin(18)/(1-2 sin(18)) definition of the golden
section
or
4 sin(18)sin(18) = 1 - 2 sin(18) = 1 - 2 cos(72)
2(1 - cos(36)) = 1 - 2 cos(72)
or
cos(36) - cos(72) = 1
Herman Serras
Herman Serras wrote:
>
> Some days ago I suggested to use a little geometry. I think following
> the same way I have a nicer proof!
of course the last line should read as
>
> cos(36) - cos(72) = 0.5
>
> Herman...@rug.ac.be
Michael Press
In article <4rtppe$l...@rzsun02.rrz.uni-hamburg.de>,
fc3...@GEOMAT.math.uni-hamburg.de (Hauke Reddmann) wrote:
> Let sin(18)=cos(72)=a,sin(36)=b,cos(36)=c,sin(72)=cos(18)=d.
> By the simplest angle-doubling formula:
>
> b=2ad c=d^2-a^2 d=2bc a=c^2-b^2 and of course a^2+d^2=b^2+c^2=1.
> So a-c=(2c^2-1)-(1-2a^2) and bd=2ad*2bc, so 1=4ac.
> Subtract to get a-c+1=2*(a-c)^2. This has solutions a-c=1,-1/2.
> As a-c is negative,it is -1/2. Q.E.D.
>
> Now for something completely different:
> cos(36)*cos(72)=1/4
cos(1x) = 1 * (cos x)
cos(2x) = 2 * (cos x)^2 - 1
cos(3x) = 4 * (cos x)^3 - 3 * cos x
cos(4x) = 8 * (cos x)^4 - 8 * (cos x)^2 + 1
cos(5x) = 16 * (cos x)^5 - 20 * (cos x)^3 + 5 * cos x
Since g := cos (18 degree) is not zero,
it is a root of 16X^4 - 20X^2 + 5.
2 * cos (72) = 20g^2 - 5 - 16g^2 + 2
= 4g^2 - 3
2 * cos (36) = 4g^2 - 2
Their difference is 1 and their product
16g^4 - 20g^2 + 6 = 1.
--
Michael Press
pre...@apple.com
thure...@gmail.com
=-2*sin54°*sin(-18°)
=2*sin54°sin18°cos18°/cos18°
=sin54°*sin36°/cos18°
=cos36°sin36°/sin72°
=2cos36°sin36°/2sin72°
=sin72°/2sin72°
=1/2
thure...@gmail.com
Vinicius Claudino Ferraz
cos x - cos (2x) = 1/2
y = cos x
y - y^2 + sin^2 x = 1/2
2y^2 - 2y^2 + 2 (1 - y^2) = 1
4y^2 - 2y - 1 = 0
y = (1 + sqrt 5)/4 or y = (1 - sqrt 5)/4
arccos y = x
twitter.com/mathspiritual
Em terça-feira, 9 de agosto de 2016 08:38:23 UTC-3, thure...@gmail.com escreveu:
rizwanrosh...@gmail.com
Cos72=sq root5-1/4
Sq root5+1/4-sqroot5-1/4=sq5-sq root5+1+1/4=1/2