Expert answers to "I have a function in the form of:cos−1(a2+bx22abx)+cos−1(c2+dx22cdx)=e"

sunyerneq

Answered question

2022-02-28

I have a function in the form of:

\cos^{- 1} (\frac{a^{2} + b x^{2}}{2 a b x}) + \cos^{- 1} (\frac{c^{2} + {d x}^{2}}{2 c d x}) = e

Chettaf04

Beginner2022-03-01Added 7 answers

u = \frac{a^{2} + b x^{2}}{2 a b x}, v = \frac{c^{2} + {d x}^{2}}{2 c d x}

to get the equation

\cos^{- 1} u + \cos^{- 1} v = e

and take the cosine of both sides:

\cos (\cos^{- 1} u + \cos^{- 1} v) = \cos e

which can be simplified in steps. First, we use Mike's suggestion to get

\cos (\cos^{- 1} u) \cos (\cos^{- 1} v) - \sin (\cos^{- 1} u) \sin (\cos^{- 1} v) = \cos e

which simplifies to

u v - \sin (\cos^{- 1} u) \sin (\cos^{- 1} v) = \sin (\cos^{- 1} v) = \cos e

but from here it is not obvious how to proceed, so we have to get tricky. The key is that

\sin (\cos^{- 1} a) = \sqrt{1 - a^{2}}

for any a, so our equation becomes

u v - \sqrt{1 - u^{2}} \sqrt{1 - v^{2}} = \cos e

which we rearrange as

\sqrt{1 - u^{2}} {1 - v^{2}} = u v - \cos e

and square both sides to get

(1 - u^{2}) (1 - v^{2}) = {(u v - \cos e)}^{2}

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