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If \(\cos(xy)\) , \(\cos x \) and \(\cos (x+y)\) are in Harmonic progression. then evaluate \(\cos x . \sec (\frac{y}{2})\)
 one year ago
 one year ago
If \(\cos(xy)\) , \(\cos x \) and \(\cos (x+y)\) are in Harmonic progression. then evaluate \(\cos x . \sec (\frac{y}{2})\)
 one year ago
 one year ago

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mathsloverBest ResponseYou've already chosen the best response.0
i tried to do like this first.
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
Should I simplify it further?
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
Or I am going on wrong track?
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
I have : \(\cfrac{2\cos (x+y) \cos (xy) }{\cos (xy) + \cos (x+y)}= \cos x\) Now : \(2\cos (x+y ) \cos (xy) = \cos x \cos (xy) + \cos x \cos (x+y)\) \(\implies 2\cos^2x \cos^2y  2\sin^2 x \sin^2 y = \cos x ( \cos (x+y) + \cos (xy))\) \(\cfrac{2(\cos^2x \cos^2y  \sin^2x \sin^2y) }{\cos(x+y) + \cos(xy)}= \cos x\) \(\cfrac{\cos^2x \cos^2y  \sin^2x \sin^2y}{\cos x\cos y }= \cos x\) \(\cos^2x\cos^2y  \sin^2x \sin^2y = \cos^2x \cos y\) \(\cos^2x \cos^2y  \cos^2 x \cos y = \sin^2x \sin^2y\) \(\cos^2x \cos^2y  \cos^2 x \cos y = (1\cos^2x)(1\cos^2y)\) \(\cos^2x\cos^2y\cos^2x\cos y = 1  \cos^2y  \cos^2x + \cos^2x\cos^2y\) \(\cancel{cos^2x\cos^2y}  \cos^2x \cos y = 1  \cos^2y  \cos^2x + \cancel{\cos^2x\cos^2y}\) \(\cos^2x\cos y = 1\cos^2y  \cos^2x\) \(1\cos^2y  \cos^2x + \cos^2x \cos y = 0 \) \((1+\cos y)(1+\cos y)  \cos^2x (1  \cos y) = 0\) \((1\cos y)(1+\cos y  \cos^2x ) = 0\) Therefore cos y = 1 and \(\cos x = \sqrt{2}\)
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
So I get :  \(\sqrt{2}\)  = \(\sqrt{2}\) @amistre64 would you please take time to check my method.
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
Any help @UnkleRhaukus ?
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
@satellite73 No clue?
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
it looks like you have done a ton of work, but i cannot do this in a few minutes, would take me at least an hour i can look at it later, it is real math not off the top of my head math
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
No problem but it didn't take me more than 10 minutes. Trust me ... its not that leanthy.
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
all though a quick google search tells me your answer is right
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
and gives a slightly shorter method here is the link but ignore it if you want http://www.qfak.com/education_reference/science_mathematics/?id=b717416#.UUsQ1BesiSo
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
great! Thanks.. A medal deserves
 one year ago
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