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PreCalc & Trig Help!
Simplify the expression using a sum or difference formula:
sin(x+y)sec x sec y
 one year ago
 one year ago
PreCalc & Trig Help! Simplify the expression using a sum or difference formula: sin(x+y)sec x sec y
 one year ago
 one year ago

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AtkinsohaBest ResponseYou've already chosen the best response.0
^opps, show all steps too please.
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
\[\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\] then multiply out
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
how do you get to that point? like, can you start from the beginning because I need to show all steps for full credit.
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
in other words, you are being asked a question to see if you know the identity multiplying out is now a matter of algebra
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
I just don't understand how you went from sin(x+y)sec(x)sec(y) to that..
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
oh i have confused you that is not the answer at all that is the identity \[\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\]
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
Yes, you have confused me..
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
now your job is to multiply \[\left(\sin(x)\cos(y)+\cos(x)\sin(y)\right)\times \sec(x)\sec(y)\]\]
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
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..
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
you get \[\sin(x)\cos(y)\sec(x)\sec(y)+\cos(x)\sin(y)\sec(x)\sec(y)\] as a first step
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
okay.. then were the heck do you go?
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
that was multiplying out using the distributive law then \(\sec(x)=\frac{1}{\cos(x)}\) so you can cancel some stuff
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
lets look at the first term \[\sin(x)\cos(y)\sec(x)\sec(y)\] that is equal to \[\frac{\sin(x)\cos(y)}{\cos(x)\cos(y)}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
cancel the \(\cos(y)\) top and bottom and get \[\frac{\sin(x)}{\cos(x)}\] which you can rewrite as \[\tan(x)\]
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
okay. that makes some sense..
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
basically the same thing for the other side too then?
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
did you know \(\sec(x)=\frac{1}{\cos(x)}\)? this is really algebra at this step
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
yes, same thing for the second term
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
only this time you will get \(\tan(y)\)
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
Yes I did. Just didn't understand the problem.. so you get down to tan x=tanx?
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
ohh, yeah. tan x + tan y
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
So is that your answer then? tan x + tan y?
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
that is my answer, yes
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
Okay, that makes sense. Do you have time to help me with one more problem?
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
okay, let me take a screen shot.
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
I kind of understand it.. but not really :/
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
ok we can do this, but it is going to take a minute ready?
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
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
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
yes, we know sin u and cos v
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
we need \(cos(u)\) do you know how to find it?
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
Don't you have to make a triangle or something?
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
Okay, I'm not sure exactly how to come about finding it on a triangle
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
dw:1362154017344:dw
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
third side via pythagoras \[\sqrt{61^211^2}=60\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
no \[\cos(u)=\frac{60}{61}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
adjacent over hypotenuse
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
Okay. so we still ned to find the sin one right?
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
right but actually i made a mistake \[\cos(u)=\frac{60}{61}\] because you are in quadrant 2
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
now we need \(\sin(v)\) and it is the same idea as before dw:1362154367573:dw
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
the third side is \[\sqrt{41^240^2}=9\]so \(\sin(v)=\frac{9}{41}\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
now plug the numbers directly in to the formula
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
okay, then you just solve it right? multiplying and adding fractions basically?
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
\[\frac{11}{61}\times (\frac{40}{41})+\frac{9}{61}\times( \frac{60}{61})\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
yeah it is arithmetic from here on in gotta run, good luck
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
oh i made a typo!!
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
for multiplying you flip the second fraction?
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
\[\frac{11}{61}\times (\frac{40}{41})+\frac{9}{41}\times( \frac{60}{61})\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.3
flip nothing multiply straight across
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
okay. ill give it a try... thanks!
 one year ago
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