Question

Please explain meaning of exp function how to use it .E.G in this (y = exp((sin^2x + sin^4 x + sin^6 x +........infinity)*ln2) ) part. ?

If ,for 0<x<pi/2 , y = exp((sin^2x + sin^4 x + sin^6 x +........infinity)*ln2) is a zero of the quadratic question x^2 - 9x +8=0 then the value of (sinx+cosx)/(sinx-cosx) is

NOTE:-This is complete question but only i have problem in using exp function E.G in y = exp((sin^2x + sin^4 x + sin^6 x +........infinity)*ln2) please explain how to use and its means since i encountered this function first time . I am okay with rest so need to solve question .please explain only exp function

Answer 1

`2 + sqrt(3)`.

Explanation:

So first we can find the zeros of the given quadratic:

`x^2-9x+8=(x-1)(x-8)` so our zeroes are `x=1,8`.

Let's now deal with the value of `y`. This requires two different tricks: log rules and a sum of an infinite geometric series.

First we see that
`y = e^(text(infinite sum) * ln(2)) = 2^(text(infinite sum))`

Then we can use the formula for an infinite sum of a geometric series:
`text(infinite sum) = sin^2(x) + sin^4(x) + ...`
`= (sin^2(x))^1 + (sin^2(x))^2 + ...`
`= (sin^2(x))/(1 - sin^2(x)) = (sin^2(x))/(cos^2(x)) = tan^2(x)`

Therefore, we know that
`2^(tan^2(x)) = 1 or 2^(tan^2(x)) = 8`
i.e.
`tan^2(x) = 0 or tan^2(x) = 3`

We can easily realize which angles these correspond to using the unit circle: `x in (0, pi/3)`. By the given constraints, `x=0` can't be the answer. Therefore, we can plug in the value and find that
`(sin(x) + cos(x))/(sin(x) - cos(x)) = (sqrt(3)+1)/(sqrt(3)-1)`

This can actually be simplified:
`(sqrt(3)+1)/(sqrt(3)-1) * (sqrt(3)+1)/(sqrt3+1) = (4 + 2sqrt(3))/2`

Hence, we either get `2 + sqrt(3)`.