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
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.