# can anyone offer advice to find ƒÆmax after one has found that cosƒÆmax=[�ã(1+15ƒÀ2)+1]/3ƒÀ ?

I do not really understand the approximation given where ƒÀ=1-ƒÃ

Thanks

Sorry

ƒÆ ----> theta

ã -----> square root

ƒÀ -----> beta

ƒÃ - ----> E

can anyone offer advice to find (theta)max after one has found that cos(theta)max=[�sqrt(1+15beta^2)+1]/3beta ?

I do not really understand the approximation given where beta=1-E

Thanks

## To find the maximum value of theta (ƒÆ), given that cos(theta)max = [�sqrt(1+15beta^2)+1]/3beta, you need to understand the approximation involved, where beta = 1 - E.

To begin, let's simplify the expression cos(theta)max = [�sqrt(1+15beta^2)+1]/3beta. Start by substituting beta = 1 - E into the equation:

cos(theta)max = [�sqrt(1 + 15(1 - E)^2) + 1]/(3(1 - E))

Now, let's break down the steps to simplify the equation further:

1. Expand the square in the 15(1 - E)^2 term:
cos(theta)max = [�sqrt(1 + 15(1 - 2E + E^2)) + 1]/(3(1 - E))
= [�sqrt(1 + 15 - 30E + 15E^2) + 1]/(3(1 - E))

2. Combine like terms inside the square root:
cos(theta)max = [�sqrt(16 + 15E^2 - 30E) + 1]/(3(1 - E))

Now, to understand the approximation given where beta = 1 - E, let's substitute that into the equation:

beta = 1 - E

This equation allows us to express E in terms of beta:

E = 1 - beta

Substituting this into the simplified expression above for cos(theta)max:

cos(theta)max = [�sqrt(16 + 15(1 - beta)^2 - 30(1 - beta)) + 1]/(3(1 - (1 - beta)))

Simplifying further:

cos(theta)max = [�sqrt(16 + 15(1 - 2beta + beta^2) - 30 + 30beta) + 1]/(3beta)

cos(theta)max = [�sqrt(16 + 15 - 30beta + 15beta^2 - 30 + 30beta) + 1]/(3beta)

cos(theta)max = [�sqrt(1 + 15beta^2) + 1]/(3beta)

Therefore, we have confirmed that the given expression cos(theta)max = [�sqrt(1 + 15beta^2) + 1]/(3beta) is the approximation provided, where beta = 1 - E.

Now that we have the expression cos(theta)max = [�sqrt(1 + 15beta^2) + 1]/(3beta), we can focus on finding the maximum value of theta (ƒÆ) by analyzing the behavior of the cosine function.

Since cos(theta) ranges between -1 and 1, the maximum value of theta occurs when cos(theta) is equal to 1.

Set cos(theta)max = 1:

1 = [�sqrt(1 + 15beta^2) + 1]/(3beta)

Multiply both sides of the equation by 3beta:

3beta = �sqrt(1 + 15beta^2) + 1

Square both sides of the equation to eliminate the square root:

9beta^2 = 1 + 15beta^2 + 2�sqrt(1 + 15beta^2) + 1

Simplify the equation:

8beta^2 + 2�sqrt(1 + 15beta^2) = 0

Now, you can solve the quadratic equation to find the value(s) of beta that satisfy this equation. Once you have the value(s) of beta, you can substitute them into beta = 1 - E to obtain the corresponding values of E. Finally, substitute the values of E into the original expression to get the maximum value(s) of theta (ƒÆ).

It's important to note that solving the quadratic equation may involve complex solutions, depending on the coefficients of the equation. Make sure to check for all possible solutions and apply the appropriate methods for finding the maximum value(s) of theta (ƒÆ).