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  1. #1
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    Trigonometry

    Prove that 2\arccos\left(\frac {11}{16}\right)+3\arccos\left(\frac 78\right)=\pi (without using a calculator...).
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by james_bond View Post
    Prove that 2\arccos\left(\frac {11}{16}\right)+3\arccos\left(\frac 78\right)=\pi (without using a calculator...).
    2\arccos\left(\frac {11}{16}\right)+3\arccos\left(\frac 78\right)

    Let's do it like a bulldozer... Take its cosine.

    -> X=\cos \left(2\arccos\left(\frac {11}{16}\right)+3\arccos\left(\frac 78\right) \right)

    We know that \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)

    --> X=\cos(2\arccos\left(\frac {11}{16}\right)) \cos(3\arccos\left(\frac 78\right)) \ \ - \ \sin(3\arccos \left(\frac 78\right) \sin(3\arccos\left(\frac 78\right)))=AB-CD

    ~~~~~~~~~~~~~~
    P. 1


    \begin{aligned} A & = \cos(2 \dots) \\<br />
& = 2 \cos^2(\dots)-1 \\<br />
& = 2 \left(\cos \left(\arccos \left(\frac{11}{16}\right) \right) \right)^2-1 \\<br />
& = 2 \left(\frac{11}{16} \right)^2-1 \\<br />
& = \boxed{\frac{-7}{128}} \end{aligned}

    ~~~~~~~~~~~~~~
    P. 2


    \begin{aligned} B & = \cos(3\arccos\left(\frac 78\right)) \\<br />
& = \cos(3 \dots) \\<br />
& = 4 \cos^3(\dots)-3 \cos(\dots) {\color{red} \ \text{demonstration available P. 5}} \\<br />
& = 4 \left(\frac 78 \right)^3 - 3 \cdot \frac 78 \\<br />
& = \boxed{\frac{7}{128}}\end{aligned}

    ~~~~~~~~~~~~~~
    P. 3



    ~~~~~~~~~~~~~~
    P. 4




    ~~~~~~~~~~~~~~
    P. 5

    \begin{aligned} \cos(3x) & =\cos(2x+x) \\<br />
& = \cos(2x)\cos(x)-\sin(2x)\sin(x) \\<br />
& = (2 \cos^2(x)-1)\cos(x)-2 \cos(x)\sin(x)\sin(x) \\<br />
& = 2 \cos^3(x)-\cos(x)-2 \cos(x)(1-\cos^2(x)) \\<br />
& = 2 \cos^3(x)-\cos(x)-2 \cos(x)+2 \cos^3(x) \\<br />
& = \boxed{4 \cos^3(x)-3 \cos(x)} \end{aligned}





    Gimme more time... Or if you want to try : \sin(\arccos(x))=\sqrt{1-x^2}
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  3. #3
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    Let \cos x = \frac{11}{16},\cos y = \frac{7}{8}
    Its already 3 in the morning... I need to sleep, so I will skip the routine computations.Maybe Moo will elaborate

    \cos(x+y) = \frac14

    \sin \frac{y}2 = \sqrt{\frac{1 - \cos y}2} = \frac14

    So:
    \cos(x+y) = \sin \frac{y}2 \Rightarrow x+y =\frac{\pi - y}2 \Rightarrow 2x + 3y = \pi

    Good night everyone
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