trig identities proof

**Wiii19** · March 23rd 2008, 04:43 PM

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**topsquark** · March 23rd 2008, 05:28 PM

Originally Posted by Wiii19

Prove: $\frac{tan^3\theta}{1+tan^2\theta} + \frac{\frac{1}{tan^3\theta}} {1+\frac{1}{tan^2\theta}} = \frac{1-2sin^2\theta cos^2\theta}{sin\theta cos\theta}$

Any help would be appreciated.

Let's simplify and add these fractions first:
$\frac{tan^3(\theta)}{1 + tan^2(\theta)} + \frac{\frac{1}{tan^3(\theta)}} {1 + \frac{1}{tan^2(\theta)}}$

$= \frac{tan^3(\theta)}{1 + tan^2(\theta)} + \frac{1} {tan^3(\theta) + tan(\theta)}$

$= \frac{tan^4(\theta) + 1}{tan(\theta)(1 + tan^2(\theta))}$

$= \frac{tan^4(\theta) + 1}{tan(\theta)~sec^2(\theta)}$

Now, when in doubt, turn everything into sines and cosines:
$= \frac{\frac{sin^4(\theta)}{cos^4(\theta)} + 1}{\frac{sin(\theta)}{cos(\theta)} \cdot \frac{1}{cos^2(\theta)}}$

$= \frac{sin^4(\theta) + cos^4(\theta)}{sin(\theta)~cos(\theta)}$

Now here's the trick. You would typically think that something like $sin^4(\theta) + cos^4(\theta)$ can't be simplified as it is the sum of two even powers. However if you do the long division:
$\frac{sin^4(\theta) + cos^4(\theta)}{sin^2(\theta) + cos^2(\theta)} = sin^2(\theta) + cos^2(\theta) - \frac{2sin^2(\theta)~cos^2(\theta)}{sin^2(\theta) + cos^2(\theta)}$

But $sin^2(\theta) + cos^2(\theta) = 1$ , so the line above really reads:
$sin^4(\theta) + cos^4(\theta) = 1 - 2sin^2(\theta)~cos^2(\theta)$

This is a nice little trick to have in your back pocket.

-Dan

**Wiii19** · March 23rd 2008, 06:15 PM

Originally Posted by topsquark

Let's simplify and add these fractions first:
$\frac{tan^3(\theta)}{1 + tan^2(\theta)} + \frac{\frac{1}{tan^3(\theta)}} {1 + \frac{1}{tan^2(\theta)}}$

$= \frac{tan^3(\theta)}{1 + tan^2(\theta)} + \frac{1} {tan^3(\theta) + tan(\theta)}$

$= \frac{tan^4(\theta) + 1}{tan(\theta)(1 + tan^2(\theta))}$

$= \frac{tan^4(\theta) + 1}{tan(\theta)~sec^2(\theta)}$

Now, when in doubt, turn everything into sines and cosines:
$= \frac{\frac{sin^4(\theta)}{cos^4(\theta)} + 1}{\frac{sin(\theta)}{cos(\theta)} \cdot \frac{1}{cos^2(\theta)}}$

$= \frac{sin^4(\theta) + cos^4(\theta)}{sin(\theta)~cos(\theta)}$

Now here's the trick. You would typically think that something like $sin^4(\theta) + cos^4(\theta)$ can't be simplified as it is the sum of two even powers. However if you do the long division:
$\frac{sin^4(\theta) + cos^4(\theta)}{sin^2(\theta) + cos^2(\theta)} = sin^2(\theta) + cos^2(\theta) - \frac{2sin^2(\theta)~cos^2(\theta)}{sin^2(\theta) + cos^2(\theta)}$

But $sin^2(\theta) + cos^2(\theta) = 1$ , so the line above really reads:
$sin^4(\theta) + cos^4(\theta) = 1 - 2sin^2(\theta)~cos^2(\theta)$

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:
$\frac{{\frac{sin^3\theta}{cos^3\theta}}} {{\frac{1}{cos^2\theta}}} + \frac{{\frac{cos^3\theta}{sin^3\theta}}} {{\frac{1}{sin^2\theta}}}$

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.

**topsquark** · March 23rd 2008, 06:25 PM

Originally Posted by Wiii19

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:
$\frac{{\frac{sin^3\theta}{cos^3\theta}}} {{\frac{1}{cos^2\theta}}} + \frac{{\frac{cos^3\theta}{sin^3\theta}}} {{\frac{1}{sin^2\theta}}}$

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.

Get rid of the complex fractions before you do anything else here. Multiply the top and bottom by cos^3(x) in the first term and the top and bottom by sin^3(x) in the second term. That will be an enormous aid in adding your fractions.

-Dan

**Wiii19** · March 23rd 2008, 06:49 PM

After this line:
$<br /> \frac{{\frac{sin^3\theta}{cos^3\theta}}} {{\frac{1}{cos^2\theta}}} + \frac{{\frac{cos^3\theta}{sin^3\theta}}} {{\frac{1}{sin^2\theta}}}<br />$

I got: $\frac{sin^3\theta cos^2\theta}{cos^3\theta} + \frac{cos^3\theta sin^2\theta}{sin^3\theta}$
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: $\frac{sin^6\theta cos^2\theta + cos^6\theta sin^2\theta}{sin^3\theta cos^3\theta}$

I don't know if that is right. If so, I just can't figure out how to go any further from here.

**topsquark** · March 23rd 2008, 06:53 PM

Originally Posted by Wiii19

After this line:
$<br /> \frac{{\frac{sin^3\theta}{cos^3\theta}}} {{\frac{1}{cos^2\theta}}} + \frac{{\frac{cos^3\theta}{sin^3\theta}}} {{\frac{1}{sin^2\theta}}}<br />$

I got: $\frac{sin^3\theta cos^2\theta}{cos^3\theta} + \frac{cos^3\theta sin^2\theta}{sin^3\theta}$

Let me demonstrate what I mean with the first term:
$\frac{\frac{sin^3(\theta)}{cos^3(\theta)}}{\frac{1 }{cos^2(\theta)}}$

$= \frac{\frac{sin^3(\theta)}{cos^3(\theta)}}{\frac{1 }{cos^2(\theta)}} \cdot \frac{cos^3(\theta)}{cos^3(\theta)}$

$= \frac{\frac{sin^3(\theta)}{cos^3(\theta)} \cdot cos^3(\theta)}{\frac{1}{cos^2(\theta)} \cdot cos^3(\theta)}$

$= \frac{sin^3(\theta)}{cos(\theta)}$

-Dan

**Wiii19** · March 23rd 2008, 07:02 PM

Oh, now I see. One last thing, do you mind explaining in detail how you got:
$\frac{sin^4(\theta) + cos^4(\theta)}{sin^2(\theta) + cos^2(\theta)} = sin^2(\theta) + cos^2(\theta) - \frac{2sin^2(\theta)~cos^2(\theta)}{sin^2(\theta) + cos^2(\theta)}$

Once again, I really appreciate the help.

**topsquark** · March 23rd 2008, 07:22 PM

Originally Posted by Wiii19

Oh, now I see. One last thing, do you mind explaining in detail how you got:
$\frac{sin^4(\theta) + cos^4(\theta)}{sin^2(\theta) + cos^2(\theta)} = sin^2(\theta) + cos^2(\theta) - \frac{2sin^2(\theta)~cos^2(\theta)}{sin^2(\theta) + cos^2(\theta)}$

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 $cos^4(x)$ term on the second line I chose to multiply $sin^2(x) + cos^2(x)$ by $cos^2(x)$ rather than the "traditional" choice of $cos^4(x) / sin^2(x)$ that we would normally use in the long division. Why? Because it worked.

-Dan

**Wiii19** · March 23rd 2008, 08:29 PM

Originally Posted by topsquark

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 $cos^4(x)$ term on the second line I chose to multiply $sin^2(x) + cos^2(x)$ by $cos^2(x)$ rather than the "traditional" choice of $cos^4(x) / sin^2(x)$ that we would normally use in the long division. Why? Because it worked.

-Dan

I was just actually thinking, can you get that same answer without doing long division or is that the only way? You can't factor or anything like that?

**topsquark** · March 23rd 2008, 09:01 PM

Originally Posted by Wiii19

I was just actually thinking, can you get that same answer without doing long division or is that the only way? You can't factor or anything like that?

There probably is another way. I just don't happen to know of any.

-Dan

**mr fantastic** · March 24th 2008, 12:03 AM

Originally Posted by topsquark

There probably is another way. I just don't happen to know of any.

-Dan

Topsquark, you're too modest!

$\frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta + \cos^2 \theta}$

$= \frac{(\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta + \cos^2 \theta}$

$= \frac{(\sin^2 \theta + \cos^2 \theta)^2}{\sin^2 \theta + \cos^2 \theta} - \frac{2 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta + \cos^2 \theta}$

$= \sin^2 \theta + \cos^2 \theta - \frac{2 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta + \cos^2 \theta}$

**Wiii19** · March 24th 2008, 08:31 AM

Originally Posted by mr fantastic

Topsquark, you're too modest!

$\frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta + \cos^2 \theta}$

$= \frac{(\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta + \cos^2 \theta}$

$= \frac{(\sin^2 \theta + \cos^2 \theta)^2}{\sin^2 \theta + \cos^2 \theta} - \frac{2 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta + \cos^2 \theta}$

$= \sin^2 \theta + \cos^2 \theta - \frac{2 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta + \cos^2 \theta}$

I don't understand how you went from $\frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta + \cos^2 \theta}$ to $\frac{(\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta + \cos^2 \theta}$

**topsquark** · March 24th 2008, 01:35 PM

Originally Posted by Wiii19

I don't understand how you went from $\frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta + \cos^2 \theta}$ to $\frac{(\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta + \cos^2 \theta}$

$sin^4(\theta) + cos^4(\theta) = sin^4(\theta) + cos^4(\theta) + 2sin^2(\theta)~cos^2(\theta) - 2sin^2(\theta)cos^2(\theta)$

$= sin^4(\theta) + 2sin^2(\theta)~cos^2(\theta) + cos^4(\theta) - 2sin^2(\theta)cos^2(\theta)$

$= (sin^2(\theta) + cos^2(\theta))^2 - 2sin^2(\theta)cos^2(\theta)$

-Dan

**topsquark** · March 24th 2008, 01:36 PM

Originally Posted by mr fantastic

Topsquark, you're too modest!

Trust me, modesty is NOT one of my cardinal sins!

-Dan

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