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Write the trigonometric expression as an algebraic expression

by jkristia
Tags: algebraic, expression, trigonometric, write
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jkristia
#1
Mar12-12, 10:35 PM
P: 54
1. The problem statement, all variables and given/known data

Not long ago I had a similar problem, which I was able to solve after reading this thread, but for this question I'm stuck and I could use a small hint.



2. Relevant equations



3. The attempt at a solution

Thanks
Attached Thumbnails
trig-question.png  
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scurty
#2
Mar12-12, 10:40 PM
P: 392
I would first solve for [itex]arctan(v)[/itex] in terms of [itex]\alpha[/itex]. See if you can do the rest!
SammyS
#3
Mar12-12, 10:55 PM
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Quote Quote by jkristia View Post
1. The problem statement, all variables and given/known data

Not long ago I had a similar problem, which I was able to solve after reading this thread, but for this question I'm stuck and I could use a small hint.



2. Relevant equations

3. The attempt at a solution

Thanks
Label the triangle differently. Label it so that the tangent of α is v .

[itex]\displaystyle \tan(\alpha)=\frac{v}{1}\,.[/itex]

jkristia
#4
Mar12-12, 11:46 PM
P: 54
Write the trigonometric expression as an algebraic expression

>>I would first solve for arctan(v) in terms of α

>>Label the triangle differently. Label it so that the tangent of α is v .

hmm, I think that is what I have, tan (α) = v = sqrt..., I'm still missing something.
scurty
#5
Mar12-12, 11:50 PM
P: 392
Quote Quote by jkristia View Post
>>I would first solve for arctan(v) in terms of α

>>Label the triangle differently. Label it so that the tangent of α is v .

hmm, I think that is what I have, tan (α) = v = sqrt..., I'm still missing something.
Okay, so you have [itex]tan(\alpha) = v[/itex]. What is [itex]arctan(v)[/itex]? Then, working from the inside out, what is [itex]sin(arctan(v))[/itex]?
jkristia
#6
Mar13-12, 08:21 AM
P: 54
That is the part I'm missing, how do I find the exact value of atan(v). I know what the result is, but I dont know how to get to it.
scurty
#7
Mar13-12, 08:35 AM
P: 392
Quote Quote by jkristia View Post
That is the part I'm missing, how do I find the exact value of atan(v). I know what the result is, but I dont know how to get to it.
Look in your original picture. v is equal to four quantities, which one would give you a nice value for [itex]arctan(v)[/itex]?
jkristia
#8
Mar13-12, 09:05 AM
P: 54
I appreciate the help and the hints, but at the moment I'm blank. I'm sure the answer is staring right at me, and it is pretty straight forward - but I just can't see it. I will think more about this during the day and hopefully tonight when I get back to it, the answer jumps right out of the picture.
jkristia
#9
Mar13-12, 09:59 AM
P: 54
ah - I think I got it, will try later. I need to find the inverse of v = sqrt()/x.
jkristia
#10
Mar13-12, 11:27 AM
P: 54
A little closer, but not there yet.
Attached Thumbnails
trig2.png  
scurty
#11
Mar13-12, 11:35 AM
P: 392
Sorry for the late reply, I took a nap!

What is [itex]arctan(tan(x))[/itex] equal to?

Can you see which value to plug in now?

Then you need to calculate sine of that number you get, which is also listed in the picture as a number not involving trig functions or trig inverses.
jkristia
#12
Mar13-12, 10:15 PM
P: 54
I have not been able to figure this out by myself, I kept going in circles. Finally when I was about to give up I Bing'ed it and found this explanation http://mathforum.org/library/drmath/view/53946.html - and now I can see how simple it is, and I can see my main problem was the labeling as was pointed out in one of the answers. Had I labeled the figure correct, then I might have been able to figure it out on my own.
scurty
#13
Mar13-12, 10:29 PM
P: 392
Yeah, I've had plenty of those moments, and then it just clicks. Just so we all know you understand it correctly, what answer did you get?
jkristia
#14
Mar14-12, 10:57 AM
P: 54
This is how I solved it. It is exactly the same steps as on the link. I read the explanation, closed the browser and solved it myself by first drawing a new trigangle with the 'correct' labels, and then it pretty much fell in place.

I think the trick is to label the triangle with the sides as v/1 = v for the inner trig function, and that is where I made the mistake.... I think



Again, thank you for your help.

Edit: I just noticed that the tanθ does not provide any additional information.
Attached Thumbnails
trig3.png  
scurty
#15
Mar14-12, 11:15 AM
P: 392
That looks good to me based off of the labels of the diagram! Just wanted to make sure you figured out the problem correctly!


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