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Trigonometric inequality on an acute triangle

Prove that in any acute triangle ABC the following inequality holds:

    \[\frac{\sin A \sin B}{\cos C} + \frac{\sin B \sin C}{\cos A} + \frac{\sin C \sin A}{\cos B} \geq \frac{9}{2}\]

Solution

Since A+B+C = \pi it holds that

(1)   \begin{equation*} \cos C = -\cos (A +B ) = -\cos A \cos B + \sin A \sin B \end{equation*}

and thus

(2)   \begin{equation*} \frac{\sin A \sin B}{\cos C} = 1 + \frac{\cos A \cos B}{\cos C} = 1 + \frac{\tan C}{\tan A + \tan B} \end{equation*}

Using Nesbitt’s inequality we see that

    \begin{align*} \sum \frac{\sin A \sin B}{\cos C} &= \sum \left ( 1 + \frac{\tan C}{\tan A + \tan B} \right ) \\ &=3 + \sum \frac{\tan C}{\tan A + \tan B} \\ &\geq 3 + \frac{3}{2} \\ &= \frac{9}{2} \end{align*}

Equality holds if-f A=B=C=\frac{\pi}{3}.

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Equality on a triangle

Let ABC be a triangle. Show that

    \[\frac{1-\cos \hat{B}}{\sin \hat{B}} \cdot \frac{1-\cos \hat{\Gamma}}{\sin \hat{\Gamma}} = 1 - \frac{2a}{a+b+c}\]

Solution

Let s denote the semiperimeter of the triangle. On account of the well known relations,

(1)   \begin{equation*}\frac{1-\cos \hat{B}}{\sin \hat{B}} = \tan \frac{\hat{B}}{2} = \sqrt{\frac{\left ( s-a \right )\left ( s-c \right )}{s\left ( s-b \right )}} \end{equation*}

(2)   \begin{equation*}\frac{1-\cos \hat{\Gamma}}{\sin \hat{\Gamma}} = \tan \frac{\hat{\Gamma}}{2} = \sqrt{\frac{\left ( s-a \right )\left ( s-b \right )}{s\left ( s-c \right )}}\end{equation*}

we have:

    \begin{align*} \frac{1-\cos \hat{B}}{\sin \hat{B}} \cdot \frac{1-\cos \hat{\Gamma}}{\sin \hat{\Gamma}} &= \tan \frac{\hat{B}}{2} \cdot \tan \frac{\hat{\Gamma}}{2} \\ &=\sqrt{\frac{\left ( s-a \right )\left ( s-c \right )}{s\left ( s-b \right )}} \cdot \sqrt{\frac{\left ( s-a \right )\left ( s-b \right )}{s\left ( s-c \right )}} \\ &=\frac{s-a}{s} \\ &= 1- \frac{2a}{a+b+c} \end{align*}

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Coincidences from … hell

Let \phi denote the golden ratio. Prove that:

    \[\sin 666^\circ = - \frac{\phi}{2}\]

Solution

First of all we note that \sin 666^\circ = - \cos 36^\circ.

 

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On the geometrical view of an integral

Evaluate the integral

    \[\mathcal{J} = \int_0^1 \sqrt{4-x^2} \; \mathrm{d}x\]

using geometric methods.

Solution

We are working on the following figure

Rendered by QuickLaTeX.com

Thus,

    \begin{align*} \int_{0}^{1} \sqrt{4-x^2} \; \mathrm{d}x &= \left ( \overset{\triangle}{\mathrm{OAB}} \right ) + \left ( \mathrm{A\overset{\frown}{O} \Gamma} \right )\\ &= \frac{\sqrt{3}}{2} + \pi \cdot 2 ^2 \cdot \frac{30}{360} \\ &= \frac{\sqrt{3}}{2} + \frac{4\pi}{12} \\ &= \frac{\pi}{3} + \frac{\sqrt{3}}{2} \end{align*}

since the red angle is 60^\circ due to the triangle \mathrm{OAB} since \tan \hat{\mathrm{O}} = \sqrt{3} ( \mathrm{OB}=1 \; , \; \mathrm{AB}=\sqrt{3} ). Therefore , the green angle is 180^\circ - 90^\circ - 60^\circ = 30^\circ. Finally, the area of the circular sector is equal to

    \[\mathrm{E} = \pi r^2 \cdot \frac{\mu}{360}\]

where \mu=30 and r=2.

 

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Trigonometric sum

Prove that

    \[\cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7}=\frac{1}{2}\]

Solution

Consider the tridiagonal matrix A =\begin{pmatrix} 1&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix} . Its eigenvalues are \displaystyle 2 \cos \frac{\pi}{7} \; , \; 2 \cos \frac{3\pi}{7} \; , \; 2 \cos \frac{5 \pi}{7}. Hence,

    \[\Tr\left ( A \right ) = 1 = 2 \left ( \cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7} \right )\]

and the result follows.

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