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Math Help - Prove that trigonometric terms are in arithmetic progression?

  1. #1
    Super Member fardeen_gen's Avatar
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    Prove that trigonometric terms are in arithmetic progression?

    If \theta_1 and \alpha are two real numbers such that \frac{\cot^4 \theta}{\csc^4 \theta}\sec^2 \alpha, \frac{1}{\csc 150^{\circ}}, \frac{\tan^4 \theta}{\sec^4 \theta}\csc^2 \alpha are in Arithmetic Progression, prove that:

    \frac{\sec^{2n} \alpha}{\sec^{2n + 2} \theta}, \frac{1}{\csc 150^{\circ}}, \frac{\csc^{2n} \alpha}{\csc^{2n + 2} \theta} are also in A.P. for all n\in \mathbb{N}
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  2. #2
    Senior Member
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    Vacuously True?

    By definition, a,b,c are in arithmetic progression iff c-b=b-a ~ a+c=2b.

    Given \frac{\cot^4 \theta}{\csc^4 \theta}\sec^2 \alpha + \frac{\tan^4 \theta}{\sec^4 \theta}\csc^2 \alpha = \frac{2}{\csc 150^{\circ}}

    Or, \frac{\sec^2 \alpha}{\sec^4\theta} + \frac{\csc^2 \alpha}{\csc^4\theta} = -\frac{4\sqrt{3}}{3}

    But the sum of two squares can never equal a negative number, therefore there are no real solutions for \theta and \alpha. Ergo, this theorem is vacuously true.
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