mathslover
  • mathslover
If \(\cos(x-y)\) , \(\cos x \) and \(\cos (x+y)\) are in Harmonic progression. then evaluate \(|\cos x . \sec (\frac{y}{2})|\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathslover
  • mathslover
i tried to do like this first.
1 Attachment
mathslover
  • mathslover
Should I simplify it further?
mathslover
  • mathslover
Or I am going on wrong track?

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mathslover
  • mathslover
I have : \(\cfrac{2\cos (x+y) \cos (x-y) }{\cos (x-y) + \cos (x+y)}= \cos x\) Now : \(2\cos (x+y ) \cos (x-y) = \cos x \cos (x-y) + \cos x \cos (x+y)\) \(\implies 2\cos^2x \cos^2y - 2\sin^2 x \sin^2 y = \cos x ( \cos (x+y) + \cos (x-y))\) \(\cfrac{2(\cos^2x \cos^2y - \sin^2x \sin^2y) }{\cos(x+y) + \cos(x-y)}= \cos x\) \(\cfrac{\cos^2x \cos^2y - \sin^2x \sin^2y}{\cos x\cos y }= \cos x\) \(\cos^2x\cos^2y - \sin^2x \sin^2y = \cos^2x \cos y\) \(\cos^2x \cos^2y - \cos^2 x \cos y = \sin^2x \sin^2y\) \(\cos^2x \cos^2y - \cos^2 x \cos y = (1-\cos^2x)(1-\cos^2y)\) \(\cos^2x\cos^2y-\cos^2x\cos y = 1 - \cos^2y - \cos^2x + \cos^2x\cos^2y\) \(\cancel{cos^2x\cos^2y} - \cos^2x \cos y = 1 - \cos^2y - \cos^2x + \cancel{\cos^2x\cos^2y}\) \(-\cos^2x\cos y = 1-\cos^2y - \cos^2x\) \(1-\cos^2y - \cos^2x + \cos^2x \cos y = 0 \) \((1+\cos y)(1+\cos y) - \cos^2x (1 - \cos y) = 0\) \((1-\cos y)(1+\cos y - \cos^2x ) = 0\) Therefore cos y = 1 and \(\cos x = \sqrt{2}\)
mathslover
  • mathslover
So I get : | \(\sqrt{2}\) | = \(\sqrt{2}\) @amistre64 would you please take time to check my method.
mathslover
  • mathslover
Any help @UnkleRhaukus ?
mathslover
  • mathslover
@satellite73 No clue?
anonymous
  • anonymous
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
mathslover
  • mathslover
No problem but it didn't take me more than 10 minutes. Trust me ... its not that leanthy.
anonymous
  • anonymous
all though a quick google search tells me your answer is right
anonymous
  • anonymous
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
mathslover
  • mathslover
great! Thanks.. A medal deserves

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