Results 1 to 6 of 6

Math Help - proving trigonometric identity

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    5

    proving trigonometric identity

    prove that sin A + cosA = 1
    secA + tanA - 1 cosecA + cotA -1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie darthfader's Avatar
    Joined
    Apr 2009
    From
    Philippines
    Posts
    10
    Quote Originally Posted by aman View Post
    prove that sin A + cosA = 1
    secA + tanA - 1 cosecA + cotA -1
    are you sure about the firs one? The \sin A + \cos A = 1?

    and the second one, what should we prove?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2009
    Posts
    5
    sorry that was mistake , sinA is divided by secA + tan A - 1 and cosA is divided by cosecA + cotA -1 and they are added to get 1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by aman View Post
    sorry that was mistake , sinA is divided by secA + tan A - 1 and cosA is divided by cosecA + cotA -1 and they are added to get 1.
    sin(A) + cos(A) \neq 1

    Try it with A = \frac{\pi}{4} = 45^o

    However: sin^2(A) + cos^2(A) = 1
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by aman View Post
    sorry that was mistake , sinA is divided by secA + tan A - 1 and cosA is divided by cosecA + cotA -1 and they are added to get 1.

    This is not an identity

    try \frac{\pi}{4}

    \sin\left( \frac{\pi}{4}\right)+\cos\left( \frac{\pi}{4}\right) =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2} \ne 1
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    TES, you've quoted the correct version of the question, but you've taken into account the very first post lol.

    We have to prove that :
    S=\frac{\sin(a)}{\sec(a)+\tan(a)-1}+\frac{\cos(a)}{\csc(a)+\cot(a)-1}=1

    Recall that \cot=\frac{\cos}{\sin}=\frac1\tan
    \sec=\frac1\cos
    \csc=\frac1\sin
    this will help you make the simplifications I'll skip


    S=\frac{\sin(a)}{\sec(a)+\tan(a)-1}=\frac{\sin(a)}{\sec(a)+\tan(a)-1}\cdot\frac{\cot(a)}{\cot(a)}=\frac{\cos(a)}{\csc  (a)+1-\cot(a)}

    Hey ! But this is very similar to \frac{\cos(a)}{\csc(a)+\cot(a)-1}


    Adding the two gives :

    \cos(a)\left(\frac{1}{\csc(a)+1-\cot(a)}+\frac{1}{\csc(a)+\cot(a)-1}\right)

    =\cos(a)\left(\frac{\csc(a)+\cot(a)-1+\csc(a)+1-\cot(a)}{(\csc(a)+1-\cot(a))(\csc(a)+\cot(a)-1)}\right)

    =\cos(a)\left(\frac{2\csc(a)}{(p-q)(p+q)}\right)

    where p=\csc(a) and q=1-\cot(a)

    \begin{aligned}<br />
(p-q)(p+q)<br />
&=p^2-q^2 \\<br />
&=\csc^2(a)-(1-\cot(a))^2 \\<br />
&=\csc^2(a)-\cot^2(a)+2\cot(a)-1<br />
\end{aligned}

    But it can be very easily proved (or known) that \csc^2(a)-\cot^2(a)=1

    Hence (p-q)(p+q)=2\cot(a)




    And finally, we have

    <br />
\frac{\sin(a)}{\sec(a)+\tan(a)-1}+\frac{\cos(a)}{\csc(a)+\cot(a)-1}=\cos(a)\cdot\frac{2\csc(a)}{2\cot(a)}

    But note that \cos\cdot\csc=\cos\cdot\frac{1}{\sin}=\cot

    Hence \boxed{\frac{\sin(a)}{\sec(a)+\tan(a)-1}+\frac{\cos(a)}{\csc(a)+\cot(a)-1}=1}



    Note : I've been doing all this while I'm not used at all to csc, cot, sec. What I used for simplifications is their definition in terms of cos, sin, tan. So the main difficulty for understanding is that I've been messy in my message and that I skipped some details lol
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving a trigonometric identity
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: May 2nd 2010, 11:35 PM
  2. Proving a Trigonometric Identity
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: December 21st 2009, 08:00 AM
  3. Proving a Trigonometric Identity
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: July 22nd 2009, 11:03 PM
  4. Proving Trigonometric Identity
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: April 25th 2009, 05:27 PM
  5. trigonometric identity proving need help
    Posted in the Trigonometry Forum
    Replies: 8
    Last Post: December 17th 2008, 07:33 AM

Search Tags


/mathhelpforum @mathhelpforum