1. ## Trigonometry

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

2. Hello,

Originally Posted by james_bond
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)) \$ $\ - \ \sin(3\arccos \left(\frac 78\right) \sin(3\arccos\left(\frac 78\right)))=AB-CD$

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P. 1

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

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P. 2

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

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P. 3

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P. 4

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P. 5

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

Gimme more time... Or if you want to try : $\sin(\arccos(x))=\sqrt{1-x^2}$

3. 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