Trig Functions in Three Problems: Graphing, Solving, and FactoringDate: 05/15/98 at 04:53:07 From: Carolyn Subject: graph sin x - cos x I am taking Math 12 by correspondence, and it's been years since I was in school. Since I haven't had math since grade 10, I'm having a few problems. I actually have three questions: 1) Sketch the graph of sin(x) - cos(x) Have I figured this out correctly? sqrt(1^2 + (-1)^2) = sqrt(2) cos(x) = 1/sqrt(2), sin(x) = -1/sqrt(2), x = - pi/4 2) Solve: sqrt(3)*cos(x) = sin(x) I don't even know where to start with this one. 3) Factor sin(x)cos^2(x) - 9 sin(x) - 9 cos(x) + sin^2(x)cos(x). I'm having a real problem with factoring. I can't seem to grasp the concept of it. It appears to me that when something is factored, parts of the original equation are just deleted. I hope that you are able to help me. Date: 05/15/98 at 08:21:17 From: Doctor Jerry Subject: Re: graph sin x - cos x Hi Carolyn, On problem 1, it looks as if you started correctly, but something happened along the way. For a*sin(x) - b*cos(x), calculate sqrt(a^2 + b^2). In your case, this is sqrt(2). Multiply and divide by it: sqrt(2)[(1/sqrt(2))*sin(x) - (1/sqrt(2))*cos(x)] Since cos(pi/4) = 1/sqrt(2) and sin(pi/4) = 1/sqrt(2), we can write: sqrt(2)[(1/sqrt(2))*sin(x) - (1/sqrt(2))*cos(x)] = sqrt(2)[cos(pi/4)*sin(x) - sin(pi/4)*cos(x)] = sqrt(2)[sin(x + pi/4)] OK, this is much easier to graph. In problem 2, you have sqrt(3)*cos(x) = sin(x). Well, you can just write: tan(x) = sqrt(3) For one thing, x = pi/3 works. In problem 3, you need to factor: sin(x)cos^2(x) - 9*sin(x) - 9*cos(x) + sin^2(x)cos(x) Let your mind look for patterns (of course, you must first have some patterns stored). From first and last terms: sin(x)*cos(x)*(cos(x) + sin(x)) From middle two: -9(sin(x) + cos(x)). So: sin(x)*cos^2(x) - 9*sin(x) - 9*cos(x) + sin^2(x)*cos(x) = sin(x)*cos(x)(cos(x) + sin(x)) - 9(sin(x) + cos(x)) = (sin(x) + cos(x))(sin(x)*cos(x) - 9) -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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