Question

Define the inverse cotangent function by restricting the domain of the cotangent function to the interval (0, π). Use a calculator to approximate the value of the expression. Round your result to two decimal places. (Enter your answer in radians.) arccot(−10)

Answer 1

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Answer 2

thank you!

Answer 3

Hmm this question is kinda tricky :[
Lemme post my thoughts and see if I have the right idea here.

Answer 4

\[\Large\bf\sf arccot(-10)\quad=\quad \theta\]\[\Large\bf\sf \cot \theta\quad=\quad -10\qquad\implies\qquad \tan \theta\quad=\quad -\frac{1}{10}\]\[\Large\bf\sf \arctan\left(-\frac{1}{10}\right)\quad=\quad \theta\]\[\Large\bf\sf \implies\qquad arccot(-10)\quad=\quad \arctan\left(-\frac{1}{10}\right)\]
I did this because most calculators have an inverse tangent button, but not an inverse cotangent button. So now we can put it into the calculator.\[\Large\bf\sf \arctan\left(-\frac{1}{10}\right)\quad\approx\quad -0.099668652\]This measure is in radians.
The output is in the 4th quadrant.
No bueno.

We are restricting our angle to the first 2 quadrants (0,π).
Tangent and Cotangent are periodic in π so we can add π to our angle to get the correct output.

\[\Large\bf\sf -0.099668652+\pi\quad\approx\quad 3.041924001~Radians\]Which is an angle between 0 and π.

Answer 5

Grrr I hope I'm doing this right :p