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Math Help - Help! Proving trigonometic identities!

  1. #1
    Junior Member StonerPenguin's Avatar
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    Help! Proving trigonometic identities!

    Ah, I've always sucked at doing proofs, so I'd like help with these two;
    Prove that sin^2 x (1 + cot^2x) = 1
    Prove that tan x (cot x + tan x) = sec^2x

    Sorry that I haven't shown my beginning steps, I'll work on it while I wait for a response. Thanks in advance for any help
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  2. #2
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    Quote Originally Posted by StonerPenguin View Post
    Ah, I've always sucked at doing proofs, so I'd like help with these two;
    Prove that sin^2 x (1 + cot^2x) = 1

    \sin^2 x (1 + \cot^2x)

    =\sin^2 x  +\sin^2 x \cot^2x

    =\sin^2 x  +\frac{\sin^2 x }{\tan^2x}

    =\sin^2 x  +\cos^2x

     <br />
=1<br />
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  3. #3
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    Quote Originally Posted by StonerPenguin View Post

    Prove that sin^2 x (1 + cot^2x) = 1
    The easiest way is to use the identity  1 \cot^2x = cosec^2x
    If you're not allowed to assume that
     LHS = \sin^2 x (1 + cot^2x)
     = \sin^2x \left(1 + \frac{\cos^2x}{\sin^2x} \right)
     = \sin^2x \left(\frac{\sin^2x + \cos^2x}{\sin^2x}\right)
    Use the identity  \sin^2x + \cos^2x = 1 and cancel out the  \sin^2x and you'll get the RHS.


    Prove that tan x (cot x + tan x) = sec^2x
     LHS = tan x (cot x + tan x)

    = \frac{\sin x}{\cos x} \left(\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} \right)

     = \frac{\sin x}{\cos x} \left(\frac{\cos^2 x + \sin^2 x}{\sin x \cos x} \right)

    Again just use  \sin^2 x + \cos^2 x = 1 and multiply out to get the RHS.
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  4. #4
    Junior Member StonerPenguin's Avatar
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    Wow! Thanks guys! Lots of smart people from Austrailia! I really appreciate the help, but Gusbob, the second one doesn't look right to me;

    Quote Originally Posted by Gusbob View Post
     LHS = tan x (cot x + tan x)

    = \frac{\sin x}{\cos x} \left(\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} \right)

     = \frac{\sin x}{\cos x} \left(\frac{\cos^2 x + \sin^2 x}{\sin x \cos x} \right)

    Again just use  \sin^2 x + \cos^2 x = 1 and multiply out to get the RHS.
    I don't see how  \frac{\sin x}{\cos x} \left(\frac{\cos^2 x + \sin^2 x}{\sin x \cos x} \right) = \sin^2 x + \cos^2 x

    Doesn't  = \frac{\sin x}{\cos x} \left(\frac{\cos^2 x + \sin^2 x}{\sin x \cos x} \right) = \frac{1}{\cos x} \left(\frac{\cos^2 x + \sin^2 x}{\cos x} \right) = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} ? And even if I'm being stupid and you are right, I'm suppose to transform the it into sec^2x not 1. (I'm so sorry if I sound too harsh or ungrateful, I just want makes sure it's right D: )

    Anywho, here's what I got, does this look right?

    \tan x (\cot x + \tan x) = \sec^2x Given
    \sec^2x = \tan x (\cot x + \tan x) Swapping the sides around
    = (\tan x \cot x)+ (\tan^2 x) Simplify by multiplying
    (\tan x \cot x) = 1 Reciprocal Identity
    = 1 + \tan^2 x = \sec^2x Pythagorean Identity
    \sec^2x = \sec^2x TAH-DAH~

    Also, here's two other question I'd like to double check;
    Find the exact value of cos 105 without using tables or a calculator.
    (Using the addition formula for cosine)
    \cos 105 = \cos (60 + 45)
    = \cos 60 \cos 45 - \sin 60 \sin 45
    = (\frac{1}{2} \left)(\frac{\sqrt2}{2} \right) - (\frac{\sqrt3}{2} \left)(\frac{\sqrt2}{2} \right)
    = \frac{\sqrt2}{4} - \frac{\sqrt6}{4}= \frac{\sqrt2 - \sqrt6}{4}

    Is that right?

    And, "Find the exact value of tan 15 without using tables or a calculator."

    (Using a Half-Angle formula for the tangent)
    \tan 15 = \tan \frac{30}{2}
    = \frac{1 - \cos 30}{\sin 30}
    {1} - \frac{\sqrt3}{2}
    _______
    .... \frac{1}{2}
    = {2} - {\sqrt3}

    BTW How do you put subordinate fractions using the math bbcode? That was the only way I could think to do it.
    Last edited by StonerPenguin; February 22nd 2010 at 08:34 AM.
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