Question

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

Answer 1

^opps, show all steps too please.

Answer 2

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

Answer 3

how do you get to that point? like, can you start from the beginning because I need to show all steps for full credit.

Answer 4

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

Answer 5

in other words, you are being asked a question to see if you know the identity
multiplying out is now a matter of algebra

Answer 6

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

Answer 7

oh i have confused you
that is not the answer at all
that is the identity
\[\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\]

Answer 8

Yes, you have confused me..

Answer 9

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

Answer 10

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..

Answer 11

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

Answer 12

okay.. then were the heck do you go?

Answer 13

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

Answer 14

lets look at the first term
\[\sin(x)\cos(y)\sec(x)\sec(y)\]
that is equal to
\[\frac{\sin(x)\cos(y)}{\cos(x)\cos(y)}\]

Answer 15

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

Answer 16

okay. that makes some sense..

Answer 17

basically the same thing for the other side too then?

Answer 18

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

Answer 19

yes, same thing for the second term

Answer 20

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

Answer 21

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

Answer 22

ohh, yeah. tan x + tan y

Answer 23

that is it, yes

Answer 24

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

Answer 25

that is my answer, yes

Answer 26

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

Answer 27

sure

Answer 28

okay, let me take a screen shot.

Answer 29

k

Answer 30

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

Answer 31

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

Answer 32

Yes.

Answer 33

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

Answer 34

yes, we know sin u and cos v

Answer 35

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

Answer 36

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

Answer 37

yes exactly

Answer 38

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

Answer 39

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

Answer 40

Wouldn't it be 50?

Answer 41

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

Answer 42

adjacent over hypotenuse

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

now plug the numbers directly in to the formula

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

thank you!

Answer 52

wait quick

Answer 53

oh i made a typo!!

Answer 54

for multiplying you flip the second fraction?

Answer 55

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

Answer 56

oh no

Answer 57

flip nothing
multiply straight across

Answer 58

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

Answer 59

yw