Jamierox4ev3r
  • 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).
Mathematics
chestercat
  • chestercat
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Jamierox4ev3r
  • Jamierox4ev3r
@ganeshie8 I actually have no idea what's going on here, so if you're online, I would love your help. thanks
ganeshie8
  • ganeshie8
recall the angle sum identity for sin : \[\sin(x+y)=\sin x\cos y + \cos x\sin y\]
ganeshie8
  • 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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Jamierox4ev3r
  • 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
  • Jamierox4ev3r
that's all we appear to know at this moment in time
ganeshie8
  • 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
  • ganeshie8
so we need to find \(\cos y\), \(\sin y\) and \(\cos x\) somehow from the given info
Jamierox4ev3r
  • Jamierox4ev3r
oh whoops, completely forgot about the secant ;-;
Jamierox4ev3r
  • Jamierox4ev3r
riiight
ganeshie8
  • ganeshie8
its easy, lets find them
Jamierox4ev3r
  • Jamierox4ev3r
Fair enough. Let's do it
ganeshie8
  • ganeshie8
do you like triangles we're gonna need bunch of them
Jamierox4ev3r
  • Jamierox4ev3r
triangles are good with me
ganeshie8
  • ganeshie8
|dw:1441088950452:dw|
ganeshie8
  • ganeshie8
try finding that missing side
Jamierox4ev3r
  • Jamierox4ev3r
that would be \(\sqrt{8}\), I believe
ganeshie8
  • ganeshie8
Yes! look at that triangle again, \(\cos x = ?\)
Jamierox4ev3r
  • Jamierox4ev3r
\(\large\frac{\sqrt{8}}{3}\)
ganeshie8
  • ganeshie8
Awesome! just draw another triangle using \(\sec y = 5/4\) and find whatever you need
ganeshie8
  • ganeshie8
maybe let me draw the triangle for you
Jamierox4ev3r
  • Jamierox4ev3r
I think I got this
ganeshie8
  • ganeshie8
good, il just wait then :)
Jamierox4ev3r
  • Jamierox4ev3r
let me try drawing it, just verify if I'm right. And thank you c:
Jamierox4ev3r
  • Jamierox4ev3r
|dw:1441089382675:dw|
Jamierox4ev3r
  • Jamierox4ev3r
oh and i forgot to draw in the right angle. but that's kind of a given
Jamierox4ev3r
  • Jamierox4ev3r
this actually works out really well, the missing side would simply be 3
ganeshie8
  • ganeshie8
yes, also the angle is \(y\), not \(x\) |dw:1441089486688:dw|
Jamierox4ev3r
  • Jamierox4ev3r
oh right, whoops . Since this time around we're solving for the y
Jamierox4ev3r
  • Jamierox4ev3r
I understand
ganeshie8
  • ganeshie8
good, find the stuff you need and see if you can finish it off!
Jamierox4ev3r
  • Jamierox4ev3r
|dw:1441089589409:dw|
Jamierox4ev3r
  • Jamierox4ev3r
sounds good to me :)
ganeshie8
  • ganeshie8
do let me know the final answer incase if you want me double check
Jamierox4ev3r
  • 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
  • Jamierox4ev3r
Can this be simplified by multiplying the fractions together?
Jamierox4ev3r
  • Jamierox4ev3r
or would this be the final form?
ganeshie8
  • 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
  • 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
  • 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
  • Jamierox4ev3r
oh wait, and isn't it possible for \(\sqrt{8}\) to be be simplified to \(2\sqrt{2}\)?
ganeshie8
  • 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
  • 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
  • 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
  • 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
  • ganeshie8
Ahh sure, ask..

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