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Trig Functions in Three Problems: Graphing, Solving, and Factoring
Date: 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
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