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iceblits
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Hey, I'm wondering if anyone knows of a trig identity for arct(a/b) where a and b are rationals.
iceblits said:Thanks for the reply. Although I can't find arctan(a/b), this is still very helpful for what I'm doing.
iceblits said:I wanted to show that arctan(a/b) may be written in the form arctan(a*m)+arctan(b*n) (or something like that) as part of a proof I am writing for a project. The entire explanation is long winded and it would take some time to explain but basically if I know that (in my project) all arctan(a) and arctan(b) and any linear combination of those exist but I have yet to show if all arctan(a/b) exist or not which is why I was hoping for a trig identity that would neatly answer the question
(aside: a/b isn't defined for all real numbers...)SteveL27 said:Well, arctan is defined for all real numbers. So no matter what a and b are, arctan(a/b) exists. Am I misunderstanding?
The formula for arctan(a/b) is arctan(a/b) = tan-1(a/b), where a and b are real numbers and b ≠ 0.
To solve for arctan(a/b) in an equation, you can use the inverse tangent function (tan-1). Plug in the values for a and b and use a calculator to find the arctan(a/b) value.
Arctan(a/b) cannot be simplified further, as it is already in its simplest form.
The range of arctan(a/b) is all real numbers between -π/2 and π/2. In other words, the output of arctan(a/b) will always be between -90° and 90°.
Arctan(a/b) is the inverse function of the tangent function, tan(a/b). In other words, arctan(a/b) "undoes" the effect of the tangent function. Additionally, arctan(a/b) can be written in terms of other trigonometric functions, such as sine and cosine: arctan(a/b) = sin-1(a/b) = cos-1(√(1-(a/b)2)).