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

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

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

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

mathslover
 one year ago
Best ResponseYou've already chosen the best response.0I 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 \(\cos x = \sqrt{2}\)

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

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

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2it 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

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2all though a quick google search tells me your answer is right

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2and 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

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