Here's the question you clicked on:
mathslover
If \(\cos(x-y)\) , \(\cos x \) and \(\cos (x+y)\) are in Harmonic progression. then evaluate \(|\cos x . \sec (\frac{y}{2})|\)
i tried to do like this first.
Should I simplify it further?
Or I am going on wrong track?
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 \(\cos x = \sqrt{2}\)
So I get : | \(\sqrt{2}\) | = \(\sqrt{2}\) @amistre64 would you please take time to check my method.
Any help @UnkleRhaukus ?
@satellite73 No clue?
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
No problem but it didn't take me more than 10 minutes. Trust me ... its not that leanthy.
all though a quick google search tells me your answer is right
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
great! Thanks.. A medal deserves