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Wilco
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16 Nov 2009, 1:02 pm

I have to solve these two equations but don't understand them at all. :S. Maybe you guys know how to do it?

sin (x-(1/4 * π) * cos x = (1/2*(2^0,5) ) * (sin x)^2

cos x * cos (4x) = sin x * sin (6x)

I fail when it comes to math XD



PlatedDrake
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16 Nov 2009, 2:34 pm

Oh, i know the process, but not the exact rules. There should be something in your text that (if memory serves cos(x)=1/(sin(x))) . . . or like that. Does the problem set have a goal (ie solve for X, or something like that)?



ruveyn
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16 Nov 2009, 2:39 pm

PlatedDrake wrote:
Oh, i know the process, but not the exact rules. There should be something in your text that (if memory serves cos(x)=1/(sin(x))) . . . or like that. Does the problem set have a goal (ie solve for X, or something like that)?


Memory does not serve. 1/sin(x) is cosec(x).

ruveyn



lau
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16 Nov 2009, 8:20 pm

Wilco wrote:
I have to solve these two equations but don't understand them at all. :S. Maybe you guys know how to do it?

sin (x-(1/4 * π) * cos x = (1/2*(2^0,5) ) * (sin x)^2

cos x * cos (4x) = sin x * sin (6x)

I fail when it comes to math XD

The first equation has mismatched parentheses, so there's no way we can guess what it is supposed to look like.

The second is pretty nasty. I might approach it from the angle that:
cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B)


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CloudWalker
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17 Nov 2009, 7:38 am

cos x * cos (4x) = sin x * sin (6x)
(cos (4x + x) + cos (4x - x)) / 2 = (cos (6x - x) - cos (6x + x)) / 2
cos 5x + cos 3x = cos 5x - cos 7x
cos 3x = -cos 7x
cos 3x + cos 7x = 0
2 cos ((7x + 3x)/2) cos ((7x - 3x)/2) = 0
cos 5x cos 2x = 0

cos 2x = 0 or cos 5x = 0
2x = 90 + 360n or 5x = 90 +360n
x = 45 + 180n or 18 + 72n



lau
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17 Nov 2009, 8:32 am

Nicely done, CloudWalker.

(I was far too out-of-practice to even start to solve it. :))


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PlatedDrake
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17 Nov 2009, 11:05 am

ruveyn wrote:
PlatedDrake wrote:
Oh, i know the process, but not the exact rules. There should be something in your text that (if memory serves cos(x)=1/(sin(x))) . . . or like that. Does the problem set have a goal (ie solve for X, or something like that)?


Memory does not serve. 1/sin(x) is cosec(x).

ruveyn


I was aiming for the general idea, not the real thing . . . but, this stuff was a bit over 6 years ago for me and i haven't kept up with it since. ><



justMax
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24 Nov 2009, 5:43 am

CloudWalker wrote:
cos x * cos (4x) = sin x * sin (6x)
(cos (4x + x) + cos (4x - x)) / 2 = (cos (6x - x) - cos (6x + x)) / 2
cos 5x + cos 3x = cos 5x - cos 7x
cos 3x = -cos 7x
cos 3x + cos 7x = 0
2 cos ((7x + 3x)/2) cos ((7x - 3x)/2) = 0
cos 5x cos 2x = 0

cos 2x = 0 or cos 5x = 0
2x = 90 + 360n or 5x = 90 +360n
x = 45 + 180n or 18 + 72n


God I envy that.

Teach me sins and cosines that well.

I just discovered Euler's Identity recently, which is odd because I can deal with partial differential equations, bra-ket notation problems, diffeomorphic manifolds, group and set theory, etc.

Yet basic trig blows my mind right now, didn't pay enough attention to really learn it beyond a brute force level, the stuff is great.



CloudWalker
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25 Nov 2009, 4:58 pm

Thanks and a late welcome too.

You must like Euler's Identity very much to use it as your avatar.



justMax
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26 Nov 2009, 2:10 pm

It is beautiful.

It shines when I consider it in my head, no stray edges, simple construction, obvious, yet still worth stating.