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sin(θ) = cos(θ) = csc(θ) = sec(θ) = cot(θ) =
I'll take a shot at it..... So tan is sin/cos which is y/x So x = 1 and y = sqrt15 Then do x^2 + y^2 = r^2
Does that seem right so far?
I have no idea. I'm been trying to figure this out for a while now. Thanks for trying! If it is right kudos to you :)
I can keep going XD
So x^2 + y^2 = r^2 (r is the hypotenuse) (1)^2 + (sqrt15)^2 = r^2 1+ 15=r^2 16=r^2 sqrt16=r r=4
So now you have all the needed numbers to finish the problem. x = 1 y = sqrt15 r = 4
sin = sqrt15/4 cos = 1/4 tan = sqrt15 csc = 4/sqrt15 sec = 4 cot = 1/sqrt15
Thank you so much! But the Sin= sqrt15/4 is not the right answer.
I can't seem to figure that one out.
Is everything else right?
Is it (sqrt15)/4 or is the 4 inside the sqrt of 15?
It is (sqrt15)/4 because it is y/r.
Ok, yeah that's the right answer. Thanks again :)
Oh okay, welcome! XD
you got it?
Yeah. But I have another problem. Use the figure and the value given below to answer the following questions. Assume A = 4. (a) Express x in terms of trigonometric ratios of θ. (b) Express y in terms of trigonometric ratios of θ.
ok...gonna need the picture tho :)
A is the Adjacent angle, y is the hypotenuse, and x is the opposite. Acute triangle.
Yes, that pic.
LIKE THIS? opps caps were on lol
The A is on the bottom, but everything else is in the right place.
are we dealing with a right triangle?
A is bottom left angle?
Just the bottom angle. I'll try to put up the pic
ahhh...thats better, when you said A was the adjacent angle it threw me :)
'sok....ill live ;)
x is equal to "sin(t)" sin(t) = x/y x = y * sin(t)
probably want my to use A for that eh....
A = 4...that helps....
I don't understand how to use it in the equation.
I think its asking for all the trig functions that use the "x" there and how we would solve for "x": x = yadayadayada stuff like that, might be overkill, but better safe than sorry :)
sin(t) = x/y ; x = y * sin(t) csc(t) = y/x ; x = y/csc(t) tan(t) = x/4 ; x = 4tan(t) cot(t) = 4/x ; x = 4/cot(t) thats all of them thatll use the x....
you know how the trig functions are defined?
What do I do for this problem though? \
the names that are given to them arent all that important, but since they got em, might as well use em. sin is simply the ration of the opposite side from the angle divided by the hypotenuse: usually abbreviated as opp/hyp.
you just express x or y in terms of the trig functions... as far as I can tell
...simply the *ratio* of the .....
oonce you know sin and cos, the rest are just made with these two...
How would I find x and y then?
you would need more information to actually solve for x and y; you would need to know the measurement of an angle, or another side so that you could work things out. But what you have is not enough information to get any "solid" answers for the rest of it.
That's all it shows in the problem.
no angle or anything.
then it cant expect you to "solve" for any values, just want you to be able to express what x and y are in terms of the rest of the information
How would I express them in terms then?
this we have enough information to solve :)
well, I wrote up all the ones I could think of for the x values up there.
if a = 4 then sin(60) = x/4 sin(60) is one of those basic angles you learn in trig class and are expected to memorize. the sin(60) = sqrt(3)/2 sqrt(3) x ----- = --- 2 4 2sqrt(3) = x is as far as you can go without a calculator
Oh.. ok I get this one. But the other has me confused :P
It just says "Express x in terms of trigonometric ratios of θ."
Same for y!
the other one just says: (a) Express x in terms of trigonometric ratios of θ. (b) Express y in terms of trigonometric ratios of θ. it doesnt say "solve" for any particular value. Just use what you know about the trig functions to be able to express x or y in terms of them
like this: sin(t) = x/y ; x = y * sin(t) csc(t) = y/x ; x = y/csc(t) tan(t) = x/4 ; x = 4tan(t) cot(t) = 4/x ; x = 4/cot(t)
How would I go about doing that? I tried the X ones you put up, but they don't seem to work.
do you have anymore options :)
then I would just skip it and move to the next question..... :)
lifes to short to worry about 1 trig problem that they dont give you enough information about right ?
Haha... Thanks so much for your help. I'm actually learning math more from you than my stupid professor lol :)
lol .... well, I do have the time.....alot of time :)
Are you in college or what?