Results 1 to 14 of 14

Math Help - trig identities proof

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    6

    Exclamation trig identities proof

    ..
    Last edited by Wiii19; March 24th 2008 at 05:20 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Wiii19 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2008
    Posts
    6

    Exclamation

    Quote Originally Posted by topsquark View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Wiii19 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2008
    Posts
    6

    Exclamation

    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Wiii19 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Mar 2008
    Posts
    6
    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Wiii19 View Post
    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
    Attached Thumbnails Attached Thumbnails trig identities proof-long-division.jpg  
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Mar 2008
    Posts
    6
    Quote Originally Posted by topsquark View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Wiii19 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by topsquark View Post
    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}
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Mar 2008
    Posts
    6
    Quote Originally Posted by mr fantastic View Post
    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}
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Wiii19 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by mr fantastic View Post
    Topsquark, you're too modest!
    Trust me, modesty is NOT one of my cardinal sins!

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. proof (compound angles) trig identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 21st 2009, 07:00 AM
  2. [SOLVED] Help with trig identities proof
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 19th 2009, 04:49 PM
  3. Trig Identities - Proof
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: November 12th 2009, 06:08 PM
  4. Proof of trig identities
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 27th 2009, 05:55 AM
  5. trig identities, proof
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: April 15th 2009, 02:26 PM

Search Tags


/mathhelpforum @mathhelpforum