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Math Help - Trigonometric Inequality

  1. #1
    Senior Member pankaj's Avatar
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    Trigonometric Inequality

    Prove that for all real x and y

     <br />
|\cos x|+|\cos y|+|\cos(x+y)| \geq 1<br />
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  2. #2
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    A few things you should know:

     \cos(x+y) = \cos x \cos y - \sin x \sin y

     |\cos x| \leq 1

     |\sin x| \leq 1
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  3. #3
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    Quote Originally Posted by abender View Post
    A few things you should know:

     \cos(x+y) = \cos x \cos y - \sin x \sin y

     |\cos x| \leq 1

     |\sin x| \leq 1
    Hello abender,

    I am curious to know how these simple identities will help. Can you give us some more hints?
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  4. #4
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    Quote Originally Posted by pankaj View Post
    Prove that for all real x and y

     <br />
|\cos x|+|\cos y|+|\cos(x+y)| \geq 1<br />
    Notice first that |\cos x| = |\cos(\pi+x)| = |\cos(\pi-x)|. So by adding multiples of pi to x and y, we may assume that both x and y lie between 0 and pi. Also, if x+y>\pi then we can replace x by \pi-x and y by \pi-y. That will have the effect of replacing x+y by 2\pi-(x+y), and again the values of |\cos x|, |\cos y| and |\cos (x+y)| will be unchanged. Therefore we can assume that x+y<\pi and hence x, y and \pi-(x+y) are the three angles of a triangle.

    So we want to prove that |\cos A| + |\cos B| + |\cos C|\geqslant1, where A, B, C are the angles of a triangle with sides a, b, c. In fact, the stronger inequality \cos A + \cos B + \cos C > 1 is true.

    The cosine rule says that \cos A + \cos B + \cos C = \dfrac{b^2+c^2-a^2}{2bc} + \dfrac{c^2+a^2-b^2}{2ca} + \dfrac{a^2+b^2-c^2}{2ab} = \dfrac{a^2(b+c) + b^2(c+a) + c^2(a+b) - a^3-b^3-c^3}{2abc}.

    So we want to prove that a^2(b+c) + b^2(c+a) + c^2(a+b) - a^3-b^3-c^3 > 2abc. But we know that b+c>a, c+a>b and a+b>c and therefore (b+c-a)(c+a-b)(a+b-c)>0. If you multiply that out, you will find that it gives exactly the inequality that is wanted.
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