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###### weird cosine question

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2 weeks ago

How would you solve the following question:

cos^x(90/x) = sqrt(0.95)

My first instinct was to use logs, but that didnt get me anywhere.

EDIT: We want the largest value of x that satisfies equation.

cos^x(90/x) = sqrt(0.95)

My first instinct was to use logs, but that didnt get me anywhere.

EDIT: We want the largest value of x that satisfies equation.

(edited 2 weeks ago)

Original post by mosaurlodon

How would you solve the following question:

cos^x(90/x) = sqrt(0.95)

My first instinct was to use logs, but that didnt get me anywhere.

EDIT: We want the largest value of x that satisfies equation.

cos^x(90/x) = sqrt(0.95)

My first instinct was to use logs, but that didnt get me anywhere.

EDIT: We want the largest value of x that satisfies equation.

Where does the question come from?

I doubt there is an exact solution but Id try solving for the smallest reciprocal so

cos(90x)^(1/x) = sqrt(0.95)

and a series expansion should get you approximately there. Though the 90 suggests working in degrees rather than radians which kinda suggests its not the way to go.

(edited 2 weeks ago)

Reply 2

2 weeks ago

https://isaacphysics.org/questions/rotating_light?stage=a_level

part C

In part B, you work out that it is I_0xcos((90)/(n))^(2n)

so for part c it is equal to 0.95*I_0

thus cos^x(90/x) = sqrt(0.95)

using desmos, you get max x as 48.1...

so minimum filters is 49.

But I am unsure how to go about solving this mathematically

part C

In part B, you work out that it is I_0xcos((90)/(n))^(2n)

so for part c it is equal to 0.95*I_0

thus cos^x(90/x) = sqrt(0.95)

using desmos, you get max x as 48.1...

so minimum filters is 49.

But I am unsure how to go about solving this mathematically

Original post by mosaurlodon

https://isaacphysics.org/questions/rotating_light?stage=a_level

part C

In part B, you work out that it is I_0xcos((90)/(n))^(2n)

so for part c it is equal to 0.95*I_0

thus cos^x(90/x) = sqrt(0.95)

using desmos, you get max x as 48.1...

so minimum filters is 49.

But I am unsure how to go about solving this mathematically

part C

In part B, you work out that it is I_0xcos((90)/(n))^(2n)

so for part c it is equal to 0.95*I_0

thus cos^x(90/x) = sqrt(0.95)

using desmos, you get max x as 48.1...

so minimum filters is 49.

But I am unsure how to go about solving this mathematically

As its an integer, the easiest seems to be just to iterate (try) different values for n to get the result or as youve done plot in desmos. Its an increasing function so just keep going. Nothing exact comes to mind and the hints are fairly poor.

Doing the previous suggestion

1 - pi^2 x / 8 + .. = sqrt(0.95)

gives 1/x = 48.7 and you could include the second quadratic term as well and solve as it may be significant or just try (it works in this case) based on a linear series expansion with may be trying a value either side. A similar thing would be to use pythagoras so cos^2n = (1-sin^2)^n and use a (linear) series approximation for sin/binomial, though if either was the way, Id have expected a hint.

(edited 2 weeks ago)

Original post by mosaurlodon

https://isaacphysics.org/questions/rotating_light?stage=a_level

part C

In part B, you work out that it is I_0xcos((90)/(n))^(2n)

so for part c it is equal to 0.95*I_0

thus cos^x(90/x) = sqrt(0.95)

using desmos, you get max x as 48.1...

so minimum filters is 49.

But I am unsure how to go about solving this mathematically

part C

In part B, you work out that it is I_0xcos((90)/(n))^(2n)

so for part c it is equal to 0.95*I_0

thus cos^x(90/x) = sqrt(0.95)

using desmos, you get max x as 48.1...

so minimum filters is 49.

But I am unsure how to go about solving this mathematically

Looks like you were correct in that there isnt a closed form solution for this part and either a graphical/sub values in or a series expansion (linear or quadratic) as in the previous posts are valid/expected ways to do it.

Reply 5

2 weeks ago

Ok thank you

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