Arrange these in order from least to greatest: arctan(-sqrt3), arctan 0, arctan(1/2) So far I got the first two values, arctan(-sqrt3), and that's 150 degrees. Arctan 0 would be zero degrees. I'll use just one answer for now, but
Let f be a function defined by f(x)= arctan x/2 + arctan x. the value of f'(0) is? It's 3/2 but I am not very clear on how to obtain the answer. I changed arctan x/2 into dy/dx=(4-2x)/(4sqrt(4+x^2)) but that's as far as I got.
Note that pi lim arctan(x ) = ---- x -> +oo 2 Now evaluate / pi \ lim |arctan(x ) - -----| x x -> +oo \ 2 / I'm not exactly sure how to attempt it. I have tried h'opital's rule but I don't believe you can use it here. Any
also: integral of tan^(-1)y dy how is integration of parts used in that? You write: arctan(y)dy = d[y arctan(y)] - y d[arctan(y)] Here we again have used the product rule: d(fg) = f dg + g df You then use that: d[arctan(y)] =
A particle moves along the x axis so that its position at any time t>= 0 is given by x = arctan t What is the limiting position of the particle as t approaches infinity? Answer is pi/2 How do I solve this? Thanks a lot. You
The values of x that are solutions to the equation cos^(2)x=sin2x in the interval [0, pi] are a. arctan(1/2) only b. arctan(1/2) and pi c. arctan(1/2) and 0 d. arctan(1/2) and (pi/2) e. arctan(1/2), o, and (pi/2)
arctan(tan(2pi/3) thanks. arctan(tan(2pi/3) = -pi/3 since arctan and tan are inverse operations, the solution would be 2pi/3 the number of solutions to arctan(x) is infinite, look at its graph. generally, unless a general solution
Now we prove Machin's formula using the tangent addition formula: tan(A+B)= tanA+tanB/1-tanAtanB. If A= arctan(120/119) and B= -arctan(1/239), how do you show that arctan(120/119)-arctan(1/239)=arctan1?
Integrate: (2x^2+5)/((x^2+1)(x^2+4)) I came up with: (tan^-1)(x)-(1/2)((tan^-1)(2/x)) but it keeps coming back the wrong answer even though I integrated correctly. Is there a way to simplify this answer, and if so, how? Your