- Jamierox4ev3r

If sin(x) = 1/3 and sec(y) = 5/4, where x and y lie between 0 and π/2, evaluate sin(x + y).

- chestercat

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- Jamierox4ev3r

@ganeshie8 I actually have no idea what's going on here, so if you're online, I would love your help. thanks

- ganeshie8

recall the angle sum identity for sin :
\[\sin(x+y)=\sin x\cos y + \cos x\sin y\]

- ganeshie8

basically you need to find those four pieces : \(\sin x, \cos x, ~~\sin y, \cos y\)
then plug them in

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## More answers

- Jamierox4ev3r

Alright, so I'm assuming that it would look a little like this:
\(\sin (x + y) = (\frac{1}{3} )(\frac{5}{4})+ \sin(y)\cos(x) \)

- Jamierox4ev3r

that's all we appear to know at this moment in time

- ganeshie8

you have the correct idea
but we don't really know the value of \(\cos y\) yet, what we're given is \(\sec y\) and not \(\cos y\)

- ganeshie8

so we need to find \(\cos y\), \(\sin y\) and \(\cos x\) somehow from the given info

- Jamierox4ev3r

oh whoops, completely forgot about the secant ;-;

- Jamierox4ev3r

riiight

- ganeshie8

its easy, lets find them

- Jamierox4ev3r

Fair enough. Let's do it

- ganeshie8

do you like triangles
we're gonna need bunch of them

- Jamierox4ev3r

triangles are good with me

- ganeshie8

|dw:1441088950452:dw|

- ganeshie8

try finding that missing side

- Jamierox4ev3r

that would be \(\sqrt{8}\), I believe

- ganeshie8

Yes!
look at that triangle again, \(\cos x = ?\)

- Jamierox4ev3r

\(\large\frac{\sqrt{8}}{3}\)

- ganeshie8

Awesome!
just draw another triangle using \(\sec y = 5/4\) and find whatever you need

- ganeshie8

maybe let me draw the triangle for you

- Jamierox4ev3r

I think I got this

- ganeshie8

good, il just wait then :)

- Jamierox4ev3r

let me try drawing it, just verify if I'm right. And thank you c:

- Jamierox4ev3r

|dw:1441089382675:dw|

- Jamierox4ev3r

oh and i forgot to draw in the right angle. but that's kind of a given

- Jamierox4ev3r

this actually works out really well, the missing side would simply be 3

- ganeshie8

yes, also the angle is \(y\), not \(x\)
|dw:1441089486688:dw|

- Jamierox4ev3r

oh right, whoops . Since this time around we're solving for the y

- Jamierox4ev3r

I understand

- ganeshie8

good, find the stuff you need and see if you can finish it off!

- Jamierox4ev3r

|dw:1441089589409:dw|

- Jamierox4ev3r

sounds good to me :)

- ganeshie8

do let me know the final answer incase if you want me double check

- Jamierox4ev3r

for sure, I want to make sure I'm doing the right thing.
So \(\sin y = \frac{3}{5}\) and \(\sin x = \frac{4}{5}\)
then we plug all of that back in to the identity
\(\large sin (x+y) = (\frac{1}{3}) (\frac{4}{5}) + (\frac{\sqrt{8}}{3}) (\frac{3}{5})\)

- Jamierox4ev3r

Can this be simplified by multiplying the fractions together?

- Jamierox4ev3r

or would this be the final form?

- ganeshie8

for sure, I want to make sure I'm doing the right thing.
So \(\sin y = \frac{3}{5}\) and \(\color{red}{\cos y }= \frac{4}{5}\)
then we plug all of that back in to the identity
\(\large sin (x+y) = (\frac{1}{3}) (\frac{4}{5}) + (\frac{\sqrt{8}}{3}) (\frac{3}{5})\)

- Jamierox4ev3r

right xD I'm sort of spacing out, but that's what I have in my notebook. Totally did not type that right

- Jamierox4ev3r

if you looked at how I plugged things into the identity, I think it illustrates that I have a good understanding now of what's going on

- Jamierox4ev3r

oh wait, and isn't it possible for \(\sqrt{8}\) to be be simplified to \(2\sqrt{2}\)?

- ganeshie8

Ahh I see it was just a typo
\(\large \sin (x+y) = (\frac{1}{3}) (\frac{4}{5}) + (\frac{\sqrt{8}}{3}) (\frac{3}{5}) = \dfrac{2(2+3\sqrt{2})}{15}\)
i would leave it at that

- Jamierox4ev3r

Yeah, that seems to be the nicest form. I got it! Thank you so much. My goodness, the calculus professor said that this was all supposed to be review. Granted, much of it is, but some of these things I feel like I've never learned before. You're a life saver, thanks a bunch!

- ganeshie8

np
btw you're doing great! I can tell, you will have no problem reviewing/learning these basic trig/functions stuff as you continue with ur calc course... just dont let these stop you from learning calc... its okay to keep coming back to these topics when you find something not so clear in ur calc course..

- Jamierox4ev3r

Yep, that's why I'm making sure I get all the review out of the way. I'm already gaining a much better understanding, so thank you very much for your efforts. I know that all my dedication to making sure I have the basics down will pay off in the long run. An thank you also for your kind words :')
It's getting pretty late for me, but I have one more question I need to ask before I leave for the day. You wouldn't mind helping me with that one as well, would you?

- ganeshie8

Ahh sure, ask..

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