anonymous
  • anonymous
Will anybody explain to me how Sine Cosine and Tangent works? I missed school when they taught us, and when they tried to explain it to me I became brutally lost. :)
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

zepdrix
  • zepdrix
Our trigonometric functions allow us to relate `the angle` of a triangle to its `sides`.
zepdrix
  • zepdrix
We have this clever acronym for remembering the relationships:\[\Large\rm \color{red}{\text{Soh}}\color{green}{\text{Cah}}\color{royalblue}{\text{Toa}}\] The `sine` of an angle is equivalent to the ratio of the `opposite` side to the `hypotenuse`. That's what the `o` and `h` stand for.\[\large\rm \color{red}{\sin x=\frac{opposite}{hypotenuse}}\]
zepdrix
  • zepdrix
Let's first look at a right triangle, and make sure we understand how to label the sides.

Looking for something else?

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

More answers

zepdrix
  • zepdrix
|dw:1444603738782:dw|So if this is my triangle, with angle x labeled here. Do you know which side to label as your `hypotenuse`?
anonymous
  • anonymous
The side that is on top of the x, aka the "long side"
zepdrix
  • zepdrix
Good. The longest side. Another way I like to think of it, is in relation to the `right angle`. It's always the side `opposite the right angle`.|dw:1444604392651:dw|
zepdrix
  • zepdrix
Hmm with that idea in mind... Which side do you think we would label as being located `opposite angle x`?
anonymous
  • anonymous
The side that is vertical. all the way to the right.
zepdrix
  • zepdrix
|dw:1444604529049:dw|
zepdrix
  • zepdrix
|dw:1444604552006:dw|Ok great. So the last side we label as being `adjacent` or `next to` or angle x.
zepdrix
  • zepdrix
|dw:1444604621344:dw|So if I have a particular triangle like this one, based on our definition of sin x, do you have an idea of how to find it using these values? :)
anonymous
  • anonymous
Inverse Cosine?
zepdrix
  • zepdrix
No no, you're getting too fancy. Using something like Inverse Sine would allow us to figure out our angle x, yes. But we weren't up to that point yet XD We just want \(\large\rm \sin x=?\)
anonymous
  • anonymous
I thought you said to find the angle :P so we want sin(x) which sine is Soa so Opposite over Adjacent?
zepdrix
  • zepdrix
Woooops :O\[\Large\rm \color{red}{\text{Soh}}\color{green}{\text{Cah}}\color{royalblue}{\text{Toa}}\]Soh, not Soa ya silly billy >.<
anonymous
  • anonymous
Oh my lord x_x How did i even pass math class... SO! It'd be Opposite over Hypotenuse, yes?
zepdrix
  • zepdrix
Ok good.\[\large\rm \color{red}{\sin x=\frac{opposite}{hypotenuse}}\]Based on the way we labeled these sides, it looks like we're using the 4 and 5, ya?\[\large\rm \color{red}{\sin x=\frac{4}{5}}\]
anonymous
  • anonymous
Yes yes. That's what the triangle tells us.
zepdrix
  • zepdrix
And ya, if we wanted to solve for the angle, we could apply the inverse sine function,\[\large\rm \arcsin\left(\sin x\right)=\arcsin\left(\frac{4}{5}\right)\]On the left you the composition of a function and it's inverse, which gives us back the argument,\[\large\rm x=\arcsin\left(\frac{4}{5}\right)\]
zepdrix
  • zepdrix
|dw:1444605278799:dw|If that's too confusing, another way to think of it is... When you change from sine to inverse sine, you just switch the stuff,
anonymous
  • anonymous
.O. ohh my... but one question... So i was told if you do normal sine, and you go to find the hypotenuse, your equasion would look something like this: 5/sin(x)= adjacent? Why is this?
anonymous
  • anonymous
*find the adjacent
zepdrix
  • zepdrix
|dw:1444605485213:dw|Ok let's see if we can do something with this triangle.
anonymous
  • anonymous
*dies* x_x okayyy umm... SOH... 4/sin(30degrees) which will give us the adjacent...
anonymous
  • anonymous
Opposite. it gives us the opposite.
zepdrix
  • zepdrix
Yah let's ignore the adjacent for now :) If we want to solve for \(\large y\),the opposite side, Then we want to pay attention to THIS stuff,|dw:1444605655180:dw|
anonymous
  • anonymous
Yes. Sin(30)= y/4
zepdrix
  • zepdrix
Mmm k good. And how would you `isolate` the y? You want to solve for y, so you need to get it alone somehow. It's being `divided` by 4 right now. How can we undo that?
anonymous
  • anonymous
multiply it by 4. which means we have to do it to the other side so 4times sin(30) = y
zepdrix
  • zepdrix
Good. And keep in mind that this 30 is like... locked in the sine function, he can't interact with the 4 in any way.\[\large\rm 4\sin(30)\ne \sin(4\cdot30)\]
zepdrix
  • zepdrix
\[\large\rm y=4\sin(30)\]And you would just use your calculator to finish that one off, unless of course you've learned about your 30/60/90 triangle, in which case you might be able to do it without a calculator.
anonymous
  • anonymous
So then we'd do sin(30) first. then multiply the answer by 4?
anonymous
  • anonymous
I got 2 when i plugged it into my calculator.
zepdrix
  • zepdrix
Yay good job! How would solve for the adjacent side, x? Let's pretend that we don't know what y is. So we can make sure of our cosine definition.
zepdrix
  • zepdrix
|dw:1444606171089:dw|Again, we don't want more variables than x. So we're dealing with these three pieces of information, ya?
anonymous
  • anonymous
Alright... so CAH... Cos(30)= x/4
anonymous
  • anonymous
and with that we'd need x alone so we multiply it by both sides correct?
anonymous
  • anonymous
4 is it XD oops.
zepdrix
  • zepdrix
\[\large\rm \cos(30)=\frac{x}{4}\]Multiply 4? Ok seems like a good idea:\[\large\rm 4\cos(30)=\frac{x}{\cancel4}\cdot\cancel4\]\[\large\rm 4\cos(30)=x\]And unfortunately, this one is going to work out to a weird decimal length, but that's ok.
anonymous
  • anonymous
I got 3.46 for x o.o
zepdrix
  • zepdrix
Here is a nice way to check your work: Recall that your hypotenuse should be the `longest side`. 3.46 is shorter than 4 \(\large\rm \color{green}{\checkmark}\) I tried to also draw that angle somewhat accurately to be a 30 degree angle. So if that vertical length is 2, would 3.5 be about right for the bottom length? yaaa that's prolly right!
anonymous
  • anonymous
Seems easy enough to remember.
zepdrix
  • zepdrix
|dw:1444606546042:dw|Let's try this problem. We need to find the length of the hypotenuse, z.
anonymous
  • anonymous
Cos(30)= 5/z
zepdrix
  • zepdrix
\[\large\rm \cos(30)=\frac{5}{z}\]Ok good. Hmmm, notice our z is in the denominator, this makes things a little trickier. Any ideas? :o
anonymous
  • anonymous
Uhhh.... divide five by both sides?...
anonymous
  • anonymous
5/cos(30)= z
zepdrix
  • zepdrix
Ya let's try that:\[\large\rm \frac{1}{5}\cdot\cos(30)=\frac{\cancel5}{z}\cdot\frac{1}{\cancel5}\]The z is still stuck in the denominator though! Ok you had the good sense to flip it after that though?\[\large\rm \frac{\cos(30)}{5}=\frac{1}{z}\qquad\to\qquad \frac{5}{\cos(30)}=z\]
zepdrix
  • zepdrix
Or maybe you got lucky :) I'm not sure which lol
anonymous
  • anonymous
XD im pretty sure i got lucky... i didn't even realize i skipped a whole piece.
zepdrix
  • zepdrix
When your variable is stuck in the denominator, these are the steps I would recommend:\[\large\rm \cos(30)=\frac{5}{z}\]Multiply both sides by z,\[\large\rm z\cos(30)=5\]Divide both sides by cos(30),\[\large\rm z=\frac{5}{\cos(30)}\]
zepdrix
  • zepdrix
So when your variable is stuck in the bottom, the process is a little bit different. Just a bit trickier.
anonymous
  • anonymous
Indeed. When I plugged the equation in, I got 3.8
zepdrix
  • zepdrix
Ok, again let's check our work. The hypotenuse should be the longest side. But our adjacent side is 5. Uh oh! 3.8 < 5
anonymous
  • anonymous
My second time i got 5.8 when i did Cos(30), got my answer, then divided 5 by my answer (5/ ans is what im trying to say...)
zepdrix
  • zepdrix
Do you have a calculator that uses the "Ans" thing?
anonymous
  • anonymous
Yes i do.
zepdrix
  • zepdrix
Ya I really like that feature. You can do the problem all at once, \(\large\rm 5\div\cos(30)\) Or in parts as you described, \(\large\rm \cos(30)\quad \boxed{=}\) \(\large\rm 5\div Ans \quad \boxed{=}\) Whichever way makes more sense to you :)
anonymous
  • anonymous
They both do. But sometimes double checking prevents me from making a mistake, gladly.
zepdrix
  • zepdrix
I need a math break I think :) lol Too much maf. Here is one more you can work on though.
zepdrix
  • zepdrix
|dw:1444607625616:dw|
zepdrix
  • zepdrix
So if you get bored and wanna try another one, try to solve for x in that triangle :)
zepdrix
  • zepdrix
I'mma go make some foods a sec >.<
anonymous
  • anonymous
Alright. I'll attempt to figure this one out. and you enjoy your food. I need a math break after this too XD I have alot more homework... lol
anonymous
  • anonymous
Tan(41)=7/x 7/tan(41)=x 7/tan(41)= 8.1 That's what I got.
anonymous
  • anonymous
Lol then i got 10.7 for the hypotenuse using the same opposite.
zepdrix
  • zepdrix
7/tan(41)=8.05, so ya 8.1 sounds good. 10.7? 0_o weird

Looking for something else?

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