Trig and Inverse Trig ProblemsDate: 30 Jul 1995 02:15:56 -0400 From: Greg Sharpe Subject: Inverse Trig, Trig Equations G'day! I have three trig-related questions. Would it be possible to include explanations for solutions and some general rules? I am using Eudora on a Macintosh and can read PICT files sent as attachments. 1) state the domain and range of y = 2 sin^-1 (2x+1) and sketch the graph (if you support PICT files) 2) if f(x) = cos^-1 (-x) + cos^-1 (x), show that f'(x) = 0 Hence determine the value of f(x) 3) solve 3sin (2x) = 2tan x, a) for 0<=x<=360 b) give the general solution Thanks, Greg. Date: 30 Jul 1995 12:36:42 -0400 From: Dr. Ken Subject: Re: Inverse Trig, Trig Equations Hello there! >1) state the domain and range of y = 2 sin^-1 (2x+1) Probably the easiest way to do this problem is to piece it together from what you already know about the Sin^-1 function. The graph of the Sin^-1 function is only defined between -1 and 1, right? That's because Sine can only output values between -1 and 1. The range of Sin^-1 is -Pi/2 to Pi/2. So let's just do this piece by piece. Here are some steps that get you from the basic Sin^-1 function to what you want: Sin^-1 (x) Sin^-1 (2x) Sin^-1 (2x + 1) 2Sin^-1 (2x + 1) Think about what each step does to change the function. For instance, to go from the first function to the second, all values of x are multiplied by 2. That means that whereas the domain of the first function was -1 to 1, the domain of the second function will be -1/2 to 1/2. Essentially, the function will be "smooshed." What effect will adding that 1 in there have? And then multiplying the whole thing by 2? I'll leave it to you to actually graph it, but if you're having trouble, please write us back and we'll give another hand. >2) if f(x) = cos^-1 (-x) + cos^-1 (x), show that f'(x) = 0 > Hence determine the value of f(x) Well, let's take the derivative of both sides. Since the derivative of Cos^-1 is -1/Sqrt{1-x^2}, we get --1/Sqrt{1-x^2} + -1/Sqrt{1-x^2}, which is zero. So yeah, the derivative of f is zero. Neat. That means f is constant. So to find the value of f, we can just plug in some number in its domain. Plugging in 0, we find that f(0) = Cos^-1 (0) + Cos^-1 (0) = Pi/2 + Pi/2 = Pi. So the f(x) = Pi everywhere. >3) solve 3sin (2x) = 2tan x, > a) for 0<=x<=360 > b) give the general solution I'll set this one up, then see if you can finish it. The first thing you should usually do (not always though, some things require creativity) is change everything into Sin(x) and Cos(x). Here we'll use the double-angle formula for Sine. Since Sin(2x) = 2Sin(x)Cos(x), we have 6Sin(x)Cos(x) = 2Sin(x)/Cos(x). Now divide through by 2Sin(x), and see what happens. -K |
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