## mathslover Group Title If $$\cos(x-y)$$ , $$\cos x$$ and $$\cos (x+y)$$ are in Harmonic progression. then evaluate $$|\cos x . \sec (\frac{y}{2})|$$ one year ago one year ago

1. mathslover

i tried to do like this first.

2. mathslover

Should I simplify it further?

3. mathslover

Or I am going on wrong track?

4. 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 $$\cos x = \sqrt{2}$$

5. mathslover

So I get : | $$\sqrt{2}$$ | = $$\sqrt{2}$$ @amistre64 would you please take time to check my method.

6. mathslover

Any help @UnkleRhaukus ?

7. mathslover

@satellite73 No clue?

8. satellite73

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

9. mathslover

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

10. satellite73