Question

\[1+\cos \alpha+\cos ^{2}\alpha+............=2-\sqrt{2}\]
find the value of alpha.

Answer 1

isn't this an infinite geometric series with r = cosx?

Answer 2

here are all the solutions
https://www.wolframalpha.com/input/?i=2+-+sqrt%282%29+%3D+1+%2F+%281+-+cos%28x%29%29

Answer 3

\(\sf{1 + \cos \alpha + \cos^2\alpha + \cos^3 \alpha + ..... = 2-\sqrt{2} \\
\cos \alpha + \cos^2 \alpha +\cos^3 \alpha + .... = 1 - \sqrt{2} \\
\cos \alpha ( 1 + \cos \alpha + \cos^2\alpha+ ....) = 1 - \sqrt{2} \\
\cos \alpha (2 - \sqrt{2} ) = 1-\sqrt{2} \\
\cos \alpha = \cfrac{1-\sqrt{2}}{2-\sqrt{2}}\\
\cos \alpha = \cfrac{(1-\sqrt{2})(2+\sqrt{2})}{2} \\
\cos \alpha = \cfrac{2 + \sqrt{2} -2\sqrt{2} - 2}{2} \\
\cos \alpha = \cfrac{-\sqrt{2}}{2} \\
}\)

Answer 4

@sourwing is right in suggesting that this is a geometric series with common ratio \[\cos \alpha\].

Looking at the given series as it now stands, I see that the initial value, a, is just 1.

The sum of a geom. series with intial value a and common ratio r is

\[S=\frac{ a }{ 1-r }\] and we are told that the sum of this particular series is \[2-\sqrt{2}\]

Answer 5

Yes mathmale. But, I am confused where I went wrong?

Answer 6

thus, you may equate S = a / (1-r) to \[2-\sqrt{2}\]

With a =1 and r = cos alpha,

\[S=\frac{ 1 }{ 1-\cos \alpha }=2-\sqrt{2}\]

Answer 7

Think you can solve this for cos alpha? for alpha alone?

Answer 8

Note that alpha must NOT be 0, since cos alpha would then be 1, and we'd have division by zero.

Answer 9

I need to hit the sack pretty pronto! Hint: I've obtained two solutions, and one of them is 3Pi/4. By all means check by substituting 3Pi/4 for alpha in the original equation for the sum of the series. Is the sum actually 2-\sqrt{2}?

Great working with you.

If you need more input from me, just private message me in the morning and I'll try to help you further. Good night.

Answer 10

By all means please try solving for alpha yourself and identifying the other root (besides 3Pi/4).

Answer 11

Okay, thanks a lot Math! :) Good Night