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Math Help - Trigonometric function

  1. #1
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    Trigonometric function

    1. By squaring both sides of the identity sin^2 A + cos^2 A=1, or otherwise, show that sin^4 A + cos^4=(1/4)(3+cos4A). Hence find the possible greatest value and the smallest value of sin^4 θ + cos^4 θ

    2. Given that A, B, C are the angles of a triangle, show that tan C= (tanA+tanB)/(tanAtanB-1)
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    Re: Trigonometric function

    Quote Originally Posted by Trefoil2727 View Post
    1. By squaring both sides of the identity sin^2 A + cos^2 A=1, or otherwise, show that sin^4 A + cos^4=(1/4)(3+cos4A). Hence find the possible greatest value and the smallest value of sin^4 θ + cos^4 θ

    2. Given that A, B, C are the angles of a triangle, show that tan C= (tanA+tanB)/(tanAtanB-1)
    What have you been able to do so far?

    -Dan
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    Re: Trigonometric function

    (sin^2 A + cos^2 A)^2 =1
    sin^4 A + 2sin^2 A cos^2 A + cos^4 A =1
    sin^4 A + cos ^4 A = 1- 2sin^2 A cos^2 A

    that's all I can do..

    For the second question, I really don't know where to start
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  4. #4
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    Re: Trigonometric function

    Quote Originally Posted by Trefoil2727 View Post
    (sin^2 A + cos^2 A)^2 =1
    sin^4 A + 2sin^2 A cos^2 A + cos^4 A =1
    sin^4 A + cos ^4 A = 1- 2sin^2 A cos^2 A

    that's all I can do..

    For the second question, I really don't know where to start
    \displaystyle \begin{align*} \sin^4{(A)} + \cos^4{(A)} &= 1 - 2\sin^2{(A)}\cos^2{(A)} \\ &= 1 - 2\left[ \sin{(A)}\cos{(A)} \right] ^2 \\ &= 1 - 2 \left[ \frac{1}{2}\sin{(2A)} \right] ^2 \\ &= 1 - 2 \cdot \frac{1}{4} \sin^2{(2A)} \\ &= 1 - \frac{1}{2} \sin^2{(2A)} \\ &= 1 - \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{2}\cos{(4A)}  \right] \\ &= 1 - \frac{1}{4} + \frac{1}{4}\cos{(4A)} \\ &= \frac{3}{4} + \frac{1}{4}\cos{(4A)} \\ &= \frac{1}{4} \left[ 3 + \cos{(4A)} \right]  \end{align*}
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    Re: Trigonometric function

    Hello, Trefoil2727!

    \text{2. Given that }A, B, C\text{ are the angles of a triangle,}
    \text{show that: }\:\tan C\:=\: \frac{\tan A+\tan B}{\tan A\tan B-1}

    \text{Since }\,A+B+C \:=\:180^o,

    . . \text{we have: }\:C \:=\:180^o - (A + B)

    \text{Then: }\;\;\tan C \;=\;\tan[180^o - (A+B)] \;=\; \frac{\tan180^o - \tan(A+B)}{1 - \tan180^o\tan(A+B)}

    . . . . . . . . . . =\;\frac{0 - \tan(A+B)}{1 - 0\tan(A+B)} \;=\;-\tan(A+B)

    . . . . . . . . . . =\;-\frac{\tan A + \tan B}{1-\tan A\tan B} \;=\;\frac{\tan A + \tan B}{\tan A\tan B - 1}
    Thanks from Trefoil2727
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