Atkinsoha
PreCalc & Trig Help!
Simplify the expression using a sum or difference formula:
sin(x+y)sec x sec y
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Atkinsoha
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^opps, show all steps too please.
anonymous
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\[\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\] then multiply out
Atkinsoha
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how do you get to that point? like, can you start from the beginning because I need to show all steps for full credit.
anonymous
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you can look in your text for a proof of that result, but that is not what you are being asked to do. you are being asked to multiply out, the point of the exercise it to use that identity, not to derive it
anonymous
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in other words, you are being asked a question to see if you know the identity
multiplying out is now a matter of algebra
Atkinsoha
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I just don't understand how you went from sin(x+y)sec(x)sec(y) to that..
anonymous
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oh i have confused you
that is not the answer at all
that is the identity
\[\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\]
Atkinsoha
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Yes, you have confused me..
anonymous
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now your job is to multiply
\[\left(\sin(x)\cos(y)+\cos(x)\sin(y)\right)\times \sec(x)\sec(y)\]\]
Atkinsoha
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Maybe I should tell you this.. I've been gone from school for a few weeks now (death in the family) I have no idea how to do any of this..
anonymous
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you get
\[\sin(x)\cos(y)\sec(x)\sec(y)+\cos(x)\sin(y)\sec(x)\sec(y)\] as a first step
Atkinsoha
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okay.. then were the heck do you go?
anonymous
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that was multiplying out using the distributive law
then \(\sec(x)=\frac{1}{\cos(x)}\) so you can cancel some stuff
anonymous
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lets look at the first term
\[\sin(x)\cos(y)\sec(x)\sec(y)\]
that is equal to
\[\frac{\sin(x)\cos(y)}{\cos(x)\cos(y)}\]
anonymous
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cancel the \(\cos(y)\) top and bottom and get
\[\frac{\sin(x)}{\cos(x)}\] which you can rewrite as
\[\tan(x)\]
Atkinsoha
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okay. that makes some sense..
Atkinsoha
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basically the same thing for the other side too then?
anonymous
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did you know \(\sec(x)=\frac{1}{\cos(x)}\)? this is really algebra at this step
anonymous
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yes, same thing for the second term
anonymous
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only this time you will get \(\tan(y)\)
Atkinsoha
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Yes I did. Just didn't understand the problem..
so you get down to tan x=tanx?
Atkinsoha
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ohh, yeah. tan x + tan y
anonymous
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that is it, yes
Atkinsoha
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So is that your answer then? tan x + tan y?
anonymous
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that is my answer, yes
Atkinsoha
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Okay, that makes sense. Do you have time to help me with one more problem?
anonymous
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sure
Atkinsoha
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okay, let me take a screen shot.
anonymous
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k
Atkinsoha
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Atkinsoha
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I kind of understand it.. but not really :/
anonymous
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ok we can do this, but it is going to take a minute
ready?
Atkinsoha
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Yes.
anonymous
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we are going to use the same formula that we used above
\[\sin(u+v)=\sin(u)\cos(v)+\cos(u)\sin(v)\] at the moment, we know only two of those four numbers
Atkinsoha
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yes, we know sin u and cos v
anonymous
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we need \(cos(u)\) do you know how to find it?
Atkinsoha
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Don't you have to make a triangle or something?
anonymous
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yes exactly
Atkinsoha
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Okay, I'm not sure exactly how to come about finding it on a triangle
anonymous
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|dw:1362154017344:dw|
anonymous
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third side via pythagoras
\[\sqrt{61^2-11^2}=60\]
Atkinsoha
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Wouldn't it be 50?
anonymous
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no
\[\cos(u)=\frac{60}{61}\]
anonymous
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adjacent over hypotenuse
Atkinsoha
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Okay. so we still ned to find the sin one right?
anonymous
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right but actually i made a mistake
\[\cos(u)=-\frac{60}{61}\] because you are in quadrant 2
anonymous
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now we need \(\sin(v)\) and it is the same idea as before |dw:1362154367573:dw|
anonymous
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the third side is
\[\sqrt{41^2-40^2}=9\]so \(\sin(v)=\frac{9}{41}\)
anonymous
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now plug the numbers directly in to the formula
Atkinsoha
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okay, then you just solve it right? multiplying and adding fractions basically?
anonymous
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\[\frac{11}{61}\times (-\frac{40}{41})+\frac{9}{61}\times(- \frac{60}{61})\]
anonymous
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yeah it is arithmetic from here on in
gotta run, good luck
Atkinsoha
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thank you!
Atkinsoha
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wait quick
anonymous
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oh i made a typo!!
Atkinsoha
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for multiplying you flip the second fraction?
anonymous
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\[\frac{11}{61}\times (-\frac{40}{41})+\frac{9}{41}\times(- \frac{60}{61})\]
anonymous
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oh no
anonymous
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flip nothing
multiply straight across
Atkinsoha
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okay. ill give it a try... thanks!
anonymous
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yw