Question

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

Answer 1

i tried to do like this first.

Answer 2

Should I simplify it further?

Answer 3

Or I am going on wrong track?

Answer 4

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 \(\cos x = \sqrt{2}\)

Answer 5

So I get : | \(\sqrt{2}\) | = \(\sqrt{2}\)

@amistre64 would you please take time to check my method.

Answer 6

Any help @UnkleRhaukus ?

Answer 7

@satellite73 No clue?

Answer 8

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

Answer 9

No problem but it didn't take me more than 10 minutes. Trust me ... its not that leanthy.

Answer 10

all though a quick google search tells me your answer is right

Answer 11

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

Answer 12

great! Thanks.. A medal deserves