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Re: angle
Posted:
Feb 11, 2005 10:36 AM
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On 10 Feb 05 15:49:36 -0500 (EST), Vladimir wrote:
>On 17 Jan 05 09:31:42 -0500 (EST), Geoff Millin wrote:
>>I have now solved the equation mentioned in my last message, (13
Jan)
>>
>>The equation was sin(x).sin(x/2-30) = sin(70-x/2).sin(100-x).
>>
>>Leaving out several intermediate steps for the sake of brevity, this
>>may be transformed into several other forms, making the answer x=80
>>sucessively more obvious:
>>
>>sin(x/2) = 2sin(3x/2-100).cos(20), which is sin(40)=2sin(20).cos(20)
>>when x=80.
>>
>>Let x/2 = y+40 so that y=0 when x=80.
>>
>>sin(y+40) = 2sin(3y+20).cos(20),
>>
>>sin(y+40) - sin(3y+40) = sin(3y), which is sin(40) - sin(40) =
sin(0)
>>when y=0.
>>
>>sin(y).[2cos(2y+40) + 4sin
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