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Math Help - Trigonometric Equations

  1. #1
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    Trigonometric Equations

    I'm not sure how to tackle this question, can anyone help me out? It asks;

    a) Show that \frac{1 - cos2x}{1 + cos2x} = tan^2x.

    b) Hence find the value of tan22.5^o in simplest exact form.
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  2. #2
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    Quote Originally Posted by Flay View Post
    I'm not sure how to tackle this question, can anyone help me out? It asks;

    a) Show that \frac{1 - cos2x}{1 + cos2x} = tan^2x.

    b) Hence find the value of tan22.5^o in simplest exact form.
    \frac{1 - cos2x}{1 + cos2x} = \frac{1 - (cos^2x - sin^2x)}{1+cos^2x-sin^2x}=\frac{1-cos^2x+sin^2x}{cos^2x+cos^2x} =\frac{sin^2x+sin^2x}{2cos^2x}=\frac{2sin^2x}{2cos  ^2x}=tan^2x

    For b, take the square root of both sides and you get an equation for tan(x), let x = 22.5.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Flay View Post
    I'm not sure how to tackle this question, can anyone help me out? It asks;

    a) Show that \frac{1 - cos2x}{1 + cos2x} = tan^2x.

    b) Hence find the value of tan22.5^o in simplest exact form.
    For part a, focus on the left side of the equation.

    First multiply the numerator and denominator by \tfrac{1}{2}

    \frac{\tfrac{1}{2}(1-\cos(2x))}{\tfrac{1}{2}(1+\cos(2x))}

    Note that \cos^2(x)=\tfrac{1}{2}(1+\cos(2x)) and \sin^2(x)=\tfrac{1}{2}(1-\cos(2x))

    Thus, the identity becomes \frac{\sin^2(x)}{\cos^2(x)}=\tan^2(x). We have proven the identity. Does this make sense?

    For part (b), note that \tan(22.5^{\circ})=\tan\left(\tfrac{45^{\circ}}{2}  \right)

    Apply the half angle identity \tan\left(\tfrac{\vartheta}{2}\right)=\frac{\sin\l  eft(\tfrac{\vartheta}{2}\right)}{\cos\left(\frac{\  vartheta}{2}\right)}=\sqrt{\frac{1-\cos\vartheta}{1+\cos\vartheta}}, where \vartheta=45^{\circ}.

    Does this make sense?

    --Chris
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  4. #4
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    Thanks, I understand now. I was making a mistake with my postive and negative signs in the simplification of the first equation.
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