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x is acute angle and
find x
Last edited by tony123 (2008-02-29 08:24:27)
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sin5x sin4x = sin3x sin8x
sin5x sin4x = sin3x 2sin4xcos4x
sin5x = 2 sin3x cos4x
sin5x = sin (3x+4x) + sin (3x - 4x)
sin5x = sin7x - sinx
sin7x - sin5x = sinx
sin (6x+x) - sin (6x-x) = sinx
2cos6xsinx = sinx
2cos6x = 1
cos6x = 0.5 = cos 60`
x = 10`
so.. x = 10 degrees!!
Was quite easy, right??
Wonder why some posts remain unanswered??
If two or more thoughts intersect, there has to be a point!
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You missed out some solutions:
Check your steps carefully again and youll see why you missed them.
In fact I can spot three errors in your proof. One of them turns out not to matter because of the condition that x is acute. The other two errors however are what led you to miss these extra solutions.
But apart from those errors, it was quite a neat proof.
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Hmmm...!!
One i've spotted!
From step 3, we can write..
sin5x sin4x - 2 sin3x sin4x cos4x = 0
sin4x (sin5x - 2 sin3x cos4x) = 0
so..
sin4x = 0 = sin 180
---> x = 45`
or
sin5x - 2sin3x cos4x = 0
---> which (as per my method) gives x = 60`
thinking about other two solutions!
Jane is too intelligent!
I shoulda been Tarzan!!
If two or more thoughts intersect, there has to be a point!
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Yes, you see
So in order to find all solutions, one has to look at cos6x = 0.5 for all possible values of 6x between 0° and 540°,
But you were brilliant in coming up with your solution. The reason nobody answered Tony123s question before is because nobody knew how to solve it.
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Got it Jane...
cos6x = 0.5 = cos60 = cos(360+60) = cos(360-60)
cos6x = cos60 = cos420 = cos300
x = 10 , 70 , 50
right?!
Thanks!
You've got a bird's eye...
If two or more thoughts intersect, there has to be a point!
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Is 0° considered an acute angle? If so, that is another solution.
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Nice point!
However, 0`=360`=2n*pi
But i think 0` is Acute..
If two or more thoughts intersect, there has to be a point!
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I take acute to refer to an angle in a triangle. You cant have an angle of 0° inside a triangle.
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