1. ## 8*cosA*cosB*CosC

Hi everyone.....
I have a problem that say:

if ( 1-8*cos(a)*cos(b)*cos(c)=0) then (a=b=c=60 degree)

**********************************

I solved it
1-8*cos(a)*cos(b)*cos(c)=

1 - 8*cos A cos B cos C =
1 - 4*cos A * [cos (B-C) + cos (B+C)]=
[sin (B-C)]^2 + [cos (B-C)]^2 - 4*cos (A) cos (B - C) + 4*[cos A]^2 =
[sin (B-C)]^2 +[cos (B-C) - 2*cos A]^2 >=0

********************************
but they said that when 1-8*cos A * cos B * cos C=0
that mean a=b=c=60 deg
?????
how can i handle this problem.?

2. ## Trigonometry

Hello blue_blue
Originally Posted by blue_blue
Hi everyone.....
I have a problem that say:

if ( 1-8*cos(a)*cos(b)*cos(c)=0) then (a=b=c=60 degree)

**********************************

I solved it
1-8*cos(a)*cos(b)*cos(c)=

1 - 8*cos A cos B cos C =
1 - 4*cos A * [cos (B-C) + cos (B+C)]=
[sin (B-C)]^2 + [cos (B-C)]^2 - 4*cos (A) cos (B - C) + 4*[cos A]^2 =
[sin (B-C)]^2 +[cos (B-C) - 2*cos A]^2 >=0

********************************
but they said that when 1-8*cos A * cos B * cos C=0
that mean a=b=c=0
?????
how can i handle this problem.?
Welcome to Math Help Forum!

I'm not sure what problem you are actually trying to solve here, but certainly $a =b=c=60^o$ is a solution to the equation. Why? Because $\cos 60^o = \tfrac12$. So if $a =b=c=60^o$,

$1-8\cos a\cos b\cos c = 1 - 8\times \tfrac12\times \tfrac12\times \tfrac12 = 1 - 1=0$

But $a=b=c=0$ is not a solution, because that gives $1-8\cos a\cos b\cos c = 1 - 8\times 1\times 1\times 1 = 1 - 8=-7$

I mean
if
1-8*cos(a)*cos(b)*cos(c)=0
that mean a=b=c=60 degree
***************************
I know that cos (60)=0.5 ==>1-8*cos 60*cos 60* cos 60=0
***************************
but i mean how can i improve that ??
**************************
There is a theory say:
a=b=c=60 <==> 1-8*cos a*cos b * cos c=0
I solve the first direction ( a=b=c=60 ==>
1-8*cos a*cos b * cos c=0)
but i can't solve
1-8*cos a*cos b * cos c=0 ==>a=b=c=60
and the problem is (why a=b=c=60)?

4. Hello blue_blue
Originally Posted by blue_blue
I mean
if
1-8*cos(a)*cos(b)*cos(c)=0
that mean a=b=c=60 degree
***************************
I know that cos (60)=0.5 ==>1-8*cos 60*cos 60* cos 60=0
***************************
but i mean how can i improve that ??
**************************
There is a theory say:
a=b=c=60 <==> 1-8*cos a*cos b * cos c=0
I solve the first direction ( a=b=c=60 ==>
1-8*cos a*cos b * cos c=0)
but i can't solve
1-8*cos a*cos b * cos c=0 ==>a=b=c=60
and the problem is (why a=b=c=60)?
Unless you know that $a = b = c$, you can't go in the other direction. There will be infinitely many solutions. For example, you could have $a = b = 0^o$ (so that $\cos a = \cos b = 1$) and $c = \arccos (0.125) = 82.82^o$. Any combination, in fact, where $\cos a\cos b\cos c = \tfrac18$.

Hello blue_blueUnless you know that $a = b = c$, you can't go in the other direction. There will be infinitely many solutions. For example, you could have $a = b = 0^o$ (so that $\cos a = \cos b = 1$) and $c = \arccos (0.125) = 82.82^o$. Any combination, in fact, where $\cos a\cos b\cos c = \tfrac18$.

THX again....
but the theory say that
1-8*cos a*cos b * cos c=0 ==>a=b=c
but a+b+c=180 deg==>a=b=c=60 deg

I solve it:
1-8*cos(a)*cos(b)*cos(c)= 0

==>1 - 8*cos A cos B cos C =0
==>1 - 4*cos A * [cos (B-C) + cos (B+C)]=0
==>[sin (B-C)]^2 + [cos (B-C)]^2 - 4*cos (A) cos (B - C) + 4*[cos A]^2 =0
==>[sin (B-C)]^2 +[cos (B-C) - 2*cos A]^2 =0
==>
[cos (B-C) - 2*cos A]^2=-[sin (B-C)]^2
==>I have 2 solution
either (first solution)
cos (B-C) - 2*cos A=-sin (B-C)
or (second solution)
cos (B-C) - 2*cos A=sin (B-C)

cos (B-C) - 2*cos A=-sin (B-C)
==>
cos (B-C)+sin(b-c)=2cos(a)

?????
but i didn't know what i'll do after that
so i can't reach that a=b=c ?

cos (B-C) - 2*cos A=sin (B-C)
==>
cos (B-C)-sin(b-c)=2cos(a)

I
i didn't know what i'll do after that too?

6. Hello blue_blue.

Ah! I see. It's a pity you mention this sooner!
Originally Posted by blue_blue
a+b+c=180 deg
I take it, then, that A, B and C are the angles of a triangle.

In that case:

$1-8\cos A\cos B\cos C = 0$

$\Rightarrow 1 - 8\cos(180-[B+C])\cos B \cos C = 0$

$\Rightarrow 1+8\cos(B+C)\cos B \cos C = 0$

$\Rightarrow 1 + 4\cos(B+C)[\cos(B+C) + \cos(B-C)]=0$

$\Rightarrow 4\cos^2(B+C)+4\cos(B-C)\cos(B+C) + 1 = 0$

This can be considered as a quadratic in $\cos(B+C)$, which has real roots only if

$16\cos^2(B-c) - 4.4.1 \ge 0$

i.e. $\cos^2(B-C) \ge 1$

But, of course, $\cos^2\theta \le 1$ for all $\theta$.

$\Rightarrow \cos^2(B-C) = 1$

which is only possible, for $0^o and $0^o, if $B = C$.

In the same way, we may eliminate $B$ and prove $A=C$.

Therefore the only solution is $A=B=C=60^o$

thanks so so so much....

8. blue_blue wrote: "if ( 1-8*cos(a)*cos(b)*cos(c)=0) then (a=b=c=60 degree)"

suggests that, 1 - 8(cos 60)(cos 60)(cos 60) = 1 - 8(1/2)(1/2)(1/2) = 0.

i) 1 - 8(cos 20)(cos 40)(cos 80) = 1 - 8(1/8) = 0?

ii) 1 - 8(cos (pi/7) cos (2pi/7) cos (3pi/7)) = 1 - 8(1/8) = 0?

beautiful isn't it?

Now i know WHY A = B = C = 60 degrees is the solution because of the restriction A + B + C = 180 degrees! THANKS
-

9. ## Re: 8*cosA*cosB*CosC

If O is the circumcentre, and H is the orthocentre, then it is well known that $(OH)^2 = R^2(1-8\cos A \cos B \cos C)$.

Then in this case, OH must be 0, and the only possible case for this is an equilateral triangle, so we are done.

Best wishes,
PowerofPoint

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