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Math Help - Trig proof involving secant and cosine

  1. #1
    Don
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    Trig proof involving secant and cosine

    The problem:

    Prove that sec^2 theta + cos^2 theta can never be less than 2.

    I can't figure out how to go about.

    I have tried to take sec^ theta as 1/cos^2 theta and cos^2 theta as sin^2 theta, but got nowhere.
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  2. #2
    Member sbhatnagar's Avatar
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    Re: Trig proof involving secant and cosine

    Quote Originally Posted by Don View Post
    The problem:

    Prove that sec^2 theta + cos^2 theta can never be less than 2.

    I can't figure out how to go about.

    I have tried to take sec^ theta as 1/cos^2 theta and cos^2 theta as sin^2 theta, but got nowhere.
    Let us take,  f(\theta)=\sec^2{ \theta} + \cos^2{\theta}.

    Now, use your knowledge of calculus to find the minimum value of f(\theta).

    You will find that minimum value of f(\theta) is 2.

    Therefore:
    f(\theta) \geq 2

    \sec^2{ \theta} + \cos^2{\theta} \geq 2

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  3. #3
    Don
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    Re: Trig proof involving secant and cosine

    Thanks for the reply.

    I am still hazy about Calculus, but I think that using algebraic identities like (a+b)^2 and (a-b)^2 this can be solved:

    sec^2 theta =1/ cos^2 theta. Therefore, cos^2 theta can be taken as 'x' and the equation reduces to x^2 + 1/x^2.

    x^2 + 1/x^2 - 2*x*1/x + 2*x*1/x.

    (x^2 - 1/x^2)^2 + 2

    This way, sec^2 theta + cos^2 theta can never be less than 2. Is this right?
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  4. #4
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    Re: Trig proof involving secant and cosine

    Yup it's correct, even if you didn't write it ordered or... well, logical.

    x^{2}+\frac{1}{x^{2}}> 2 \Leftrightarrow x^{2}+\frac{1}{x^{2}}-2>0 \Leftrightarrow x^{2}+\frac{1}{x^{2}}-2\cdot x^{2}\cdot\frac{1}{x^{2}}>0 \Leftrightarrow \left ( x-\frac{1}{x} \right )^{2}>0
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