Write the trigonometric expression as an algebraic expression

[jkristia]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

Thanks

 

[scurty]

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

 

[SammyS]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

Thanks

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

$\displaystyle \tan(\alpha)=\frac{v}{1}\,.$

 

[jkristia]

>>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]

>>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 $tan(\alpha) = v$. What is $arctan(v)$? Then, working from the inside out, what is $sin(arctan(v))$?

 

[jkristia]

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 don't know how to get to it.

 

[scurty]

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 don't 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 $arctan(v)$?

 

[jkristia]

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]

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

 

[jkristia]

A little closer, but not there yet.

 

[scurty]

Sorry for the late reply, I took a nap!

What is $arctan(tan(x))$ 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]

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]

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]

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.

 

[scurty]

That looks good to me based off of the labels of the diagram! Just wanted to make sure you figured out the problem correctly!