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

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

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

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

^opps, show all steps too please.

- anonymous

\[\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\] then multiply out

- 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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

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

I just don't understand how you went from sin(x+y)sec(x)sec(y) to that..

- 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

Yes, you have confused me..

- anonymous

now your job is to multiply
\[\left(\sin(x)\cos(y)+\cos(x)\sin(y)\right)\times \sec(x)\sec(y)\]\]

- 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

you get
\[\sin(x)\cos(y)\sec(x)\sec(y)+\cos(x)\sin(y)\sec(x)\sec(y)\] as a first step

- anonymous

okay.. then were the heck do you go?

- anonymous

that was multiplying out using the distributive law
then \(\sec(x)=\frac{1}{\cos(x)}\) so you can cancel some stuff

- 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

cancel the \(\cos(y)\) top and bottom and get
\[\frac{\sin(x)}{\cos(x)}\] which you can rewrite as
\[\tan(x)\]

- anonymous

okay. that makes some sense..

- anonymous

basically the same thing for the other side too then?

- anonymous

did you know \(\sec(x)=\frac{1}{\cos(x)}\)? this is really algebra at this step

- anonymous

yes, same thing for the second term

- anonymous

only this time you will get \(\tan(y)\)

- anonymous

Yes I did. Just didn't understand the problem..
so you get down to tan x=tanx?

- anonymous

ohh, yeah. tan x + tan y

- anonymous

that is it, yes

- anonymous

So is that your answer then? tan x + tan y?

- anonymous

that is my answer, yes

- anonymous

Okay, that makes sense. Do you have time to help me with one more problem?

- anonymous

sure

- anonymous

okay, let me take a screen shot.

- anonymous

k

- anonymous

##### 1 Attachment

- anonymous

I kind of understand it.. but not really :/

- anonymous

ok we can do this, but it is going to take a minute
ready?

- anonymous

Yes.

- 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

yes, we know sin u and cos v

- anonymous

we need \(cos(u)\) do you know how to find it?

- anonymous

Don't you have to make a triangle or something?

- anonymous

yes exactly

- anonymous

Okay, I'm not sure exactly how to come about finding it on a triangle

- anonymous

|dw:1362154017344:dw|

- anonymous

third side via pythagoras
\[\sqrt{61^2-11^2}=60\]

- anonymous

Wouldn't it be 50?

- anonymous

no
\[\cos(u)=\frac{60}{61}\]

- anonymous

adjacent over hypotenuse

- anonymous

Okay. so we still ned to find the sin one right?

- anonymous

right but actually i made a mistake
\[\cos(u)=-\frac{60}{61}\] because you are in quadrant 2

- anonymous

now we need \(\sin(v)\) and it is the same idea as before |dw:1362154367573:dw|

- anonymous

the third side is
\[\sqrt{41^2-40^2}=9\]so \(\sin(v)=\frac{9}{41}\)

- anonymous

now plug the numbers directly in to the formula

- anonymous

okay, then you just solve it right? multiplying and adding fractions basically?

- anonymous

\[\frac{11}{61}\times (-\frac{40}{41})+\frac{9}{61}\times(- \frac{60}{61})\]

- anonymous

yeah it is arithmetic from here on in
gotta run, good luck

- anonymous

thank you!

- anonymous

wait quick

- anonymous

oh i made a typo!!

- anonymous

for multiplying you flip the second fraction?

- anonymous

\[\frac{11}{61}\times (-\frac{40}{41})+\frac{9}{41}\times(- \frac{60}{61})\]

- anonymous

oh no

- anonymous

flip nothing
multiply straight across

- anonymous

okay. ill give it a try... thanks!

- anonymous

yw

Looking for something else?

Not the answer you are looking for? Search for more explanations.