anonymous
  • anonymous
PreCalc & Trig Help! Simplify the expression using a sum or difference formula: sin(x+y)sec x sec y
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
^opps, show all steps too please.
anonymous
  • anonymous
\[\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\] then multiply out
anonymous
  • anonymous
how do you get to that point? like, can you start from the beginning because I need to show all steps for full credit.

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

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