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

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Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0^opps, show all steps too please.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3\[\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\] then multiply out

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3you 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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3in other words, you are being asked a question to see if you know the identity multiplying out is now a matter of algebra

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3oh 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
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, you have confused me..

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3now your job is to multiply \[\left(\sin(x)\cos(y)+\cos(x)\sin(y)\right)\times \sec(x)\sec(y)\]\]

Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0Maybe 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..

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3you get \[\sin(x)\cos(y)\sec(x)\sec(y)+\cos(x)\sin(y)\sec(x)\sec(y)\] as a first step

Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0okay.. then were the heck do you go?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3that was multiplying out using the distributive law then \(\sec(x)=\frac{1}{\cos(x)}\) so you can cancel some stuff

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3lets look at the first term \[\sin(x)\cos(y)\sec(x)\sec(y)\] that is equal to \[\frac{\sin(x)\cos(y)}{\cos(x)\cos(y)}\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3cancel the \(\cos(y)\) top and bottom and get \[\frac{\sin(x)}{\cos(x)}\] which you can rewrite as \[\tan(x)\]

Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0okay. that makes some sense..

Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0basically the same thing for the other side too then?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3did you know \(\sec(x)=\frac{1}{\cos(x)}\)? this is really algebra at this step

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3yes, same thing for the second term

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3only this time you will get \(\tan(y)\)

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

Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0ohh, yeah. tan x + tan y

Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0So is that your answer then? tan x + tan y?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3that is my answer, yes

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

Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0okay, let me take a screen shot.

Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0I kind of understand it.. but not really :/

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3ok we can do this, but it is going to take a minute ready?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3we 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
 2 years ago
Best ResponseYou've already chosen the best response.0yes, we know sin u and cos v

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3we need \(cos(u)\) do you know how to find it?

Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0Don't you have to make a triangle or something?

Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, I'm not sure exactly how to come about finding it on a triangle

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3dw:1362154017344:dw

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3third side via pythagoras \[\sqrt{61^211^2}=60\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3no \[\cos(u)=\frac{60}{61}\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3adjacent over hypotenuse

Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0Okay. so we still ned to find the sin one right?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3right but actually i made a mistake \[\cos(u)=\frac{60}{61}\] because you are in quadrant 2

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3now we need \(\sin(v)\) and it is the same idea as before dw:1362154367573:dw

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3the third side is \[\sqrt{41^240^2}=9\]so \(\sin(v)=\frac{9}{41}\)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3now plug the numbers directly in to the formula

Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0okay, then you just solve it right? multiplying and adding fractions basically?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3\[\frac{11}{61}\times (\frac{40}{41})+\frac{9}{61}\times( \frac{60}{61})\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3yeah it is arithmetic from here on in gotta run, good luck

Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0for multiplying you flip the second fraction?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3\[\frac{11}{61}\times (\frac{40}{41})+\frac{9}{41}\times( \frac{60}{61})\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3flip nothing multiply straight across

Atkinsoha
 2 years ago
Best ResponseYou've already chosen the best response.0okay. ill give it a try... thanks!
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