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The latter is not a deduction from the former. It is simply a definition, which the author is free to make, as the symbol B' has not been used up to that point.Terrell said:
yes. i got that part, but how do i get from the 2nd equation to the third equation?blue_leaf77 said:
but how do i deduce it? what is step1? i don't know how to start.. from A'x'^2 + B'x'y' + ... + F = 0 to B' = 2(C - A)...andrewkirk said:
This thread is six years old. Don't count on a response from the participants.mkeaudric said:
Cotangent is a trigonometric function that represents the ratio of the adjacent side to the opposite side of a right triangle.
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. In this case, cot2(theta) can be written as (A-C)/B, which follows the same format as the Pythagorean theorem (c^2 = a^2 + b^2).
In this equation, A, B, and C represent the lengths of the sides of a right triangle, with A being the adjacent side, B being the opposite side, and C being the hypotenuse.
To solve for theta, you would use inverse trigonometric functions. In this case, you would take the inverse cotangent (arccot) of both sides of the equation. This would give you theta = arccot((A-C)/B).
Yes, cot2(theta) can be negative. Just like with any other trigonometric function, the value of cot2(theta) can be positive or negative depending on the position of the angle theta in the coordinate plane.
