Wednesday

August 20, 2014

August 20, 2014

Posted by **kiran** on Wednesday, July 7, 2010 at 9:03pm.

- trig -
**Reiny**, Wednesday, July 7, 2010 at 10:36pmare we solving ?

cotxcot2x-cot2xcot3x-cot3xcotx=1

(cosx/sinx)(cos2x/sin2x) - (cos2x/sin2x)(cos3x/sin3x) - (cos3x/sin3x)(cosx/sinx) = 1

common denominator is sinx(sin2x)(sin3x), so

[ cosxcos2xsin3x - sinxcos2xcos3x - sin2xcos3xcosx ] / (sinxsin2xsin3x) = 1

cross-multiply, then bring sinxsin2xsin3x back to left side

cosxcos2xsin3x - sinxcos2xcos3x - sin2xcos3xcosx - sinxsin2xsin3x = 0

factor sin3x from 1st and 4th, cos3 from 2nd and 3rd

sin3x(cosxcos2x - sinxsin2x) - cos3x(sinxcos2x + cosxsin2x) = 0

sin3x(cos(x+2x)) - cos3x(sin(x+2x)) = 0

sin3xcos3x - cos3xsin3x = 0

0 = 0

Since this is a true statement, the original equation must have been an identity.

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