funky question part 2 (1 Viewer)

[kwabon]

assuming cos 3x = 4(cos x)^3 - 3cos x, where cos 3x = 1/2
show 8y^3 - 6y - 1 = 0, where y = cos x (DONE)
hence evaluate cos (pi/9) cos (2pi/9) cos (4pi/9) (DONE BUT VERY DODGEY)
my question is, i looked at the answers and it said
line 1; cos (pi/9) cos (5pi/9) cos (7pi/9) = 1/8
line 2; cos (pi/9) - cos (4pi/9) - cos (2pi/9) = 1/8
WTF at line 2, they minused 180 from it, instead of the usual 360, can you even do that? could someone go about doing this question and please explain the evaluation bit, because i dont get whats happening.
thanks.

 

[Trebla]

assuming cos 3x = 4(cos x)^3 - 3cos x, where cos 3x = 1/2
show 8y^3 - 6y - 1 = 0, where y = cos x (DONE)
hence evaluate cos (pi/9) cos (2pi/9) cos (4pi/9) (DONE BUT VERY DODGEY)
my question is, i looked at the answers and it said
line 1; cos (pi/9) cos (5pi/9) cos (7pi/9) = 1/8
line 2; cos (pi/9) - cos (4pi/9) - cos (2pi/9) = 1/8
WTF at line 2, they minused 180 from it, instead of the usual 360, can you even do that? could someone go about doing this question and please explain the evaluation bit, because i dont get whats happening.
thanks.

Note that:
cos(π - x) = - cos x
which is where the negative comes from

 

[inamotoichiban]

cos(pi - x) = -cosx
so cos5pi/9 = -cos4pi/9
and cos7pi/9 = -cos2pi/9

so when you times them together the negatives cancel out and it all works out

 

[kwabon]

Note that:
cos(π - x) = - cos x
which is where the negative comes from

so you can do that right? you dont always have to minus it or add by 2pi always right?

 

[Trebla]

Depends on what result you want...

 

[untouchablecuz]

so you can do that right? you dont always have to minus it or add by 2pi always right?

look at it in terms of the quadrants (ATSC)
<br>cos(x) is negative when in the second quad .'. cos(π-x)=-cosx<br>cos(x) is negative when in the third quad .'. cos(π+x)=-cosx<br>cos(x) is positive in the third quad .'. cos(2π-x)=cos(π)<br><br>do the same with sine and tangent functions<br>sin(x) positive in second quad .'. sin(π-x)=sin(x)<br>sin(x) is negative in third quad .'. sin(π+x)=-sin(x)<br>sin(x) is negative in fourth quad .'. sin(2π-x)=-sin(x)

 

[Bunzhou]

what paper is this from? looks very familiar

 

[cutemouse]

Looks like a CSSA question to me.