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Let's simplify and add these fractions first:
$\displaystyle \frac{tan^3(\theta)}{1 + tan^2(\theta)} + \frac{\frac{1}{tan^3(\theta)}} {1 + \frac{1}{tan^2(\theta)}}$
$\displaystyle = \frac{tan^3(\theta)}{1 + tan^2(\theta)} + \frac{1} {tan^3(\theta) + tan(\theta)}$
$\displaystyle = \frac{tan^4(\theta) + 1}{tan(\theta)(1 + tan^2(\theta))}$
$\displaystyle = \frac{tan^4(\theta) + 1}{tan(\theta)~sec^2(\theta)}$
Now, when in doubt, turn everything into sines and cosines:
$\displaystyle = \frac{\frac{sin^4(\theta)}{cos^4(\theta)} + 1}{\frac{sin(\theta)}{cos(\theta)} \cdot \frac{1}{cos^2(\theta)}}$
$\displaystyle = \frac{sin^4(\theta) + cos^4(\theta)}{sin(\theta)~cos(\theta)}$
Now here's the trick. You would typically think that something like $\displaystyle sin^4(\theta) + cos^4(\theta)$ can't be simplified as it is the sum of two even powers. However if you do the long division:
$\displaystyle \frac{sin^4(\theta) + cos^4(\theta)}{sin^2(\theta) + cos^2(\theta)} = sin^2(\theta) + cos^2(\theta) - \frac{2sin^2(\theta)~cos^2(\theta)}{sin^2(\theta) + cos^2(\theta)}$
But $\displaystyle sin^2(\theta) + cos^2(\theta) = 1$, so the line above really reads:
$\displaystyle sin^4(\theta) + cos^4(\theta) = 1 - 2sin^2(\theta)~cos^2(\theta)$
This is a nice little trick to have in your back pocket.
-Dan
Your method is much more efficient, but in the beginning I first changed all terms into sines and cosines. So a couple of lines down, after the first line, it reads:
$\displaystyle \frac{{\frac{sin^3\theta}{cos^3\theta}}} {{\frac{1}{cos^2\theta}}} + \frac{{\frac{cos^3\theta}{sin^3\theta}}} {{\frac{1}{sin^2\theta}}}$
The problem that I'm having is combining those 2 fractions. Should I multiply them by their reciprocal? If so, after I do that, I would have to make the bases the same and that's were it gets a little confusing. So, if you don't mind explaining how to get the answer from the line I gave or is that a lost cause? Thanks in advance.
After this line:
$\displaystyle
\frac{{\frac{sin^3\theta}{cos^3\theta}}} {{\frac{1}{cos^2\theta}}} + \frac{{\frac{cos^3\theta}{sin^3\theta}}} {{\frac{1}{sin^2\theta}}}
$
I got: $\displaystyle \frac{sin^3\theta cos^2\theta}{cos^3\theta} + \frac{cos^3\theta sin^2\theta}{sin^3\theta}$
Then I tried to make the bases the same by multiplying the first side by sin^3x then the other side by cos^3x. This is where I'm really confused, I got: $\displaystyle \frac{sin^6\theta cos^2\theta + cos^6\theta sin^2\theta}{sin^3\theta cos^3\theta}$
I don't know if that is right. If so, I just can't figure out how to go any further from here.
Let me demonstrate what I mean with the first term:
$\displaystyle \frac{\frac{sin^3(\theta)}{cos^3(\theta)}}{\frac{1 }{cos^2(\theta)}}$
$\displaystyle = \frac{\frac{sin^3(\theta)}{cos^3(\theta)}}{\frac{1 }{cos^2(\theta)}} \cdot \frac{cos^3(\theta)}{cos^3(\theta)}$
$\displaystyle = \frac{\frac{sin^3(\theta)}{cos^3(\theta)} \cdot cos^3(\theta)}{\frac{1}{cos^2(\theta)} \cdot cos^3(\theta)}$
$\displaystyle = \frac{sin^3(\theta)}{cos(\theta)}$
-Dan
Oh, now I see. One last thing, do you mind explaining in detail how you got:
$\displaystyle \frac{sin^4(\theta) + cos^4(\theta)}{sin^2(\theta) + cos^2(\theta)} = sin^2(\theta) + cos^2(\theta) - \frac{2sin^2(\theta)~cos^2(\theta)}{sin^2(\theta) + cos^2(\theta)}$
Once again, I really appreciate the help.
Oh dear. I hate typing these things out in the forum, so I hope this works.
This should speak for itself except for one non-standard step. When I was working to cancel the $\displaystyle cos^4(x)$ term on the second line I chose to multiply $\displaystyle sin^2(x) + cos^2(x)$ by $\displaystyle cos^2(x)$ rather than the "traditional" choice of $\displaystyle cos^4(x) / sin^2(x)$ that we would normally use in the long division. Why? Because it worked.
-Dan
Topsquark, you're too modest!
$\displaystyle \frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta + \cos^2 \theta}$
$\displaystyle = \frac{(\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta + \cos^2 \theta}$
$\displaystyle = \frac{(\sin^2 \theta + \cos^2 \theta)^2}{\sin^2 \theta + \cos^2 \theta} - \frac{2 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta + \cos^2 \theta}$
$\displaystyle = \sin^2 \theta + \cos^2 \theta - \frac{2 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta + \cos^2 \theta}$
$\displaystyle sin^4(\theta) + cos^4(\theta) = sin^4(\theta) + cos^4(\theta) + 2sin^2(\theta)~cos^2(\theta) - 2sin^2(\theta)cos^2(\theta)$
$\displaystyle = sin^4(\theta) + 2sin^2(\theta)~cos^2(\theta) + cos^4(\theta) - 2sin^2(\theta)cos^2(\theta)$
$\displaystyle = (sin^2(\theta) + cos^2(\theta))^2 - 2sin^2(\theta)cos^2(\theta)$
-Dan