My Math Forum A very hard inverse trig problem

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 October 2nd, 2007, 01:46 PM #1 Newbie   Joined: Oct 2007 From: waterloo Posts: 2 Thanks: 0 A very hard inverse trig problem Hi everyone. I cannot be able to solve the question below. This is a slightly modified problem comes from the Stewart's Calculus Early Transcendentals, under the trig section. If anyone has an idea or solution to this question please show or explain. Thank you very much. Question: A pendulum swings back and forth. Its horizontal position x (in meters) at time t (in seconds) is given by: x(t) = 1 + 2cos[pi/3(t+1)] Q1: Find a formula for the inverse function t=t(x) when the domain of x(t) is restricted so that t is greater or equal to -1 and less or equal to 2. Express the answer in the form of A Arccos (B(x-a)) +b. Q2: Find a formula for the inverse function t=t(x) when the domain of x(t) is restricted so that t is greater or equal to 2 and less or equal to 5. Express the answer in the form of C Arcsin (D(x-c)) +d.
 October 2nd, 2007, 06:39 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,739 Thanks: 1810 Q1. x(t) = 1 + 2cos[pi/3(t+1)] (where -1 ≤ t ≤ 2) implies t = (3/pi)arccos((x - 1)/2) - 1. Did you get that far? For Q2, use the identity cos(y) ≡ sin(y - 3pi/2).

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# Hardest question in inverse function

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