Write the trigonometric expression as an algebraic expressionby jkristia Tags: algebraic, expression, trigonometric, write 

#1
Mar1212, 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 



#2
Mar1212, 10:40 PM

P: 379

I would first solve for [itex]arctan(v)[/itex] in terms of [itex]\alpha[/itex]. See if you can do the rest!




#3
Mar1212, 10:55 PM

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



#4
Mar1212, 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. 



#5
Mar1212, 11:50 PM

P: 379





#6
Mar1312, 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.




#7
Mar1312, 08:35 AM

P: 379





#8
Mar1312, 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.




#9
Mar1312, 09:59 AM

P: 54

ah  I think I got it, will try later. I need to find the inverse of v = sqrt()/x.




#10
Mar1312, 11:27 AM

P: 54

A little closer, but not there yet.




#11
Mar1312, 11:35 AM

P: 379

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. 



#12
Mar1312, 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.




#13
Mar1312, 10:29 PM

P: 379

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?




#14
Mar1412, 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. 



#15
Mar1412, 11:15 AM

P: 379

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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