I need help with I just can't seem to get anywhere. this is as far as I have got:
Solve for b
arcsin(b)+ 2arctan(b)=pi
arcsin(b)=pi-2arctan(b)
b=sin(pi-2arctan(b))
Sub in Sin difference identity
let 2U=(2arctan(b))
sin(a-b)=sinacosb-cosasinb
=(sin(pi))(cos(2U))-(cos(pi))(sin(2U))
=(0)(cos(2U))-(-1)(sin(2U))
=(sin(2u))
b=sin(2arctan(b))
Now what should I do?
sin(2arctan(b))=
2 sin(arctan(b))cos(arctanb)
If you draw a right triangle with the lenghts of the two sides at right angles b and 1, so that one of the angeles becomes arctan(b), you see that
sin(arctan(b)) = b/sqrt[1+b^2]
cos(arctan(b)) = 1/sqrt[1+b^2]
To solve for b in the equation sin(2arctan(b)) = 2 sin(arctan(b)) cos(arctan(b)), you can substitute the values of sin(arctan(b)) and cos(arctan(b)) as derived before.
sin(arctan(b)) = b/sqrt(1+b^2)
cos(arctan(b)) = 1/sqrt(1+b^2)
Now, substitute these values into the equation:
sin(2arctan(b)) = 2 sin(arctan(b)) cos(arctan(b))
sin(2arctan(b)) = 2 * (b/sqrt(1+b^2)) * (1/sqrt(1+b^2))
To simplify further, multiply the two terms inside the parentheses:
sin(2arctan(b)) = (2b) / (sqrt(1+b^2) * sqrt(1+b^2))
sin(2arctan(b)) = (2b) / (1+b^2)
Now, set this expression equal to the given value of pi:
(2b) / (1+b^2) = pi
You now have a single equation (2b) / (1+b^2) = pi that you can solve to find the value of b. To simplify this equation further, you can cross-multiply to get:
2b = pi(1+b^2)
Expand the right side:
2b = pi + pi*b^2
Rearrange the equation to isolate the quadratic term:
pi*b^2 - 2b + pi = 0
This is a quadratic equation in terms of b. To solve for b, you can use the quadratic formula:
b = (-(-2) ± sqrt((-2)^2 - 4*pi*pi)) / (2*pi)
Simplifying further,
b = (2 ± sqrt(4 + 4*pi^2)) / (2*pi)
b = (1 ± sqrt(1 + pi^2)) / pi
Now you have the solution for b in terms of pi.