# Trigonometry Set-6

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Example - 14

Let $$S$$ be an interior point of triangle $$ABC.$$ Show that at least one of \begin{align}\angle SAB,\ \angle SBC,\ \text{and}\ \angle SCA\end{align} is less than or equal to 30º.

Solution: The given conditions in the problem motivate us to consider the Brocard point $$P$$ of triangle \begin{align}ABC\ \text{with}\ \alpha =\angle PAB=\angle PBC=\angle PCA\end{align}. Because S (see figure) lies inside or on the boundary of at least one of the triangles $$PAB, PBC,$$ and $$PCA,$$ at least one of \begin{align}\angle SAB,\ \angle SBC,\ \text{and}\ \angle SCA\end{align} is less than or equal to $$\alpha$$. It suffices to show that \begin{align}\alpha \le 30{}^\text{o}\end{align}; that is, \begin{align}\sin \alpha \le \frac{1}{2}\ \text{or}\ {{\csc }^{2}}\alpha \ge 4\end{align}, by considering the range of $$\alpha$$.

We have shown that

${{\csc }^{2}}\alpha ={{\csc }^{2}}A+{{\csc }^{2}}B+{{\csc }^{2}}C$

and using the Cauchy-Schwarz inequality, we have

\begin{align}\frac{9}{4}{\csc ^2}\alpha \ge \left( {{{\sin }^2}A + {{\sin }^2}B + {{\sin }^2}C} \right)\left( {{{\csc }^2}A + {{\csc }^2}B + {{\csc }^2}C} \right) \ge 9,\end{align}

implying that \begin{align}{\csc ^2}\alpha \ge 4\end{align}, as desired.

Example - 15

Let $$x, y, z$$ be positive real numbers.

(a) Prove that

\begin{align} \frac{x}{{\sqrt {1 + {x^2}} }} + \frac{y}{{\sqrt {1 + {y^2}} }} + \frac{z}{{\sqrt {1 + {z^2}} }} \le \frac{{3\sqrt 3 }}{2}\\ {\text{if}}\;x + y + z = xyz; \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \end{align}

(b) Prove that

\begin{align} \frac{x}{{1 - {x^2}}} + \frac{y}{{1 - {y^2}}} + \frac{z}{{1 - {z^2}}} \ge \frac{{3\sqrt 3 }}{2}\\ {\text{if}}\;0 < x,\;y,\;z < 1\;{\rm{and}}\;xy + yz + zx = 1. \qquad \qquad \qquad \qquad \qquad \end{align}

Solution: Both problems can be solved by trigonometric substitutions.

(a) Let there be there is an acute triangle $$ABC$$ with

$$\tan A = x,\;\tan B = y$$, and $$\tan C = z$$. Note that

\begin{align}\frac{{\tan A}}{{\sqrt {1 + {{\tan }^2}A} }} = \frac{{\tan A}}{{\sec A}} = \sin A.\end{align}

The desired inequality becomes

\begin{align}\sin A + \sin B + \sin C \le \frac{{3\sqrt 3 }}{2}\end{align}

which we’ve covered earlier.

(b) From the given condition, we can assume that there is an acute triangle $$ABC$$ such that

\begin{align}\tan \frac{A}{2} = x, \qquad \tan \frac{B}{2} = y, \qquad \tan \frac{C}{2} = z.\end{align}

By the double-angle formulas, it suffices to prove that

\begin{align}\tan A + \tan B + \tan C \ge 3\sqrt 3 \end{align}

which is another inequality we’ve encountered already.

Example - 16

Compute the sums

$\left( \begin{matrix} n \\ 1 \\ \end{matrix} \right)\sin a+\left( \begin{matrix} n \\ 2 \\ \end{matrix} \right)\sin 2a+....+\left( \begin{matrix} n \\ n \\ \end{matrix} \right)\sin na$

and

$\left( \begin{matrix} n \\ 1 \\ \end{matrix} \right)\cos a+\left( \begin{matrix} n \\ 2 \\ \end{matrix} \right)\cos 2a+....+\left( \begin{matrix} n \\ n \\ \end{matrix} \right)\cos na$

Solution: Let Sn and Tn denote the first and second sums, respectively. Set the complex number $$z=\cos a+i\sin a$$. Then, by de Moivre’s formula, we have \begin{align}{{z}^{n}}=\cos na+i\sin na\end{align}. By the binomial theorem, we obtain.

\begin{align} 1+{{T}_{n}}+i\,{{S}_{n}}& =1+\left( \begin{matrix} n \\ 1 \\ \end{matrix} \right)(\cos a+i\ \sin a)+\left( \begin{matrix} n \\ 2 \\ \end{matrix} \right)(\cos 2a+i\ \sin 2a) \\ & \quad \;\;\;\; +...+\left( \begin{matrix} n \\ n \\ \end{matrix} \right)(\cos na+i\ \sin na) \\ & =\left( \begin{matrix} n \\ 0 \\ \end{matrix} \right){{z}^{0}}+\left( \begin{matrix} n \\ 1 \\ \end{matrix} \right)z+\left( \begin{matrix} n \\ 2 \\ \end{matrix} \right) {{z}^{2}}+....+\left( \begin{matrix} n \\ n \\ \end{matrix} \right){{z}^{n}} \\ & ={{(1+z)}^{n}} \\ \end{align}

Because

\begin{align} 1 + z = 1 + \cos a + i\;\sin a & = 2{\cos ^2}\frac{a}{2} + 2i\sin \frac{a}{2}\cos \frac{a}{2}\\ & = 2\cos \frac{a}{2}\left( {\cos \frac{a}{2} + i\;\sin \frac{a}{2}} \right), \end{align}

if follows that

\begin{align}{(1 + z)^n} = {2^n}\cos \frac{a}{2}\left( {\cos \frac{{na}}{2} + i\;\sin \frac{{na}}{2}} \right),\end{align}

again by de Moivre’s formula. Therefore,

\begin{align}(1 + {T_n}) + i\,{S_n} = \left( {{2^n}{{\cos }^n}\frac{a}{2}\cos \frac{{na}}{2}} \right) + i\,\left( {{2^n}{{\cos }^n}\frac{a}{2}\sin \frac{{na}}{2}} \right),\end{align}

and so

\begin{align}{S_n} = {2^n}{\cos ^n}\frac{a}{2}\sin \frac{{na}}{2}\;{\rm{and}}\;{T_n} = - 1 + {2^n}{\cos ^n}\frac{a}{2}\cos \frac{{na}}{2}.\end{align}

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