Question

Question number 1,2,3,4,5 on
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Answer 1

Answer on Question # 56499 – Math – Trigonometry

1. If $AB$ is a chord of contact of point $P(5, -5)$ to the circle $x^2 + y^2 = 5$ and $(\alpha, \beta)$ is the orthocenter of $\Delta PAB$, then $|\alpha - 1| + |\beta - 1|$ is equal to

(A) 2 (B) $\sqrt{5}$ (C) $2\sqrt{5}$ (D) 3

Solution

The radius of the circle is $\sqrt{5}$.

Define the coordinates of the point tangent to the circle. Radius is the diagonal of a square: $2a^2 = 5$, $a = \sqrt{\frac{5}{2}}$

The point tangent to the circle $A\left(-\sqrt{\frac{5}{2}}, -\sqrt{\frac{5}{2}}\right)$.

The point tangent to the circle $B\left(\sqrt{\frac{5}{2}}, \sqrt{\frac{5}{2}}\right)$.

The equation of the line $AP: \frac{x - 5}{-\sqrt{\frac{5}{2}} - 5} = \frac{y + 5}{-\sqrt{\frac{5}{2}} + 5}; \left(5 - \sqrt{\frac{5}{2}}\right)x + \left(5 + \sqrt{\frac{5}{2}}\right)y = -10\sqrt{\frac{5}{2}}$.

Find the equation of height drawn from vertex B:

$$\frac{x - \sqrt{\frac{5}{2}}}{5 - \sqrt{\frac{5}{2}}} = \frac{y - \sqrt{\frac{5}{2}}}{5 + \sqrt{\frac{5}{2}}}; \left(5 + \sqrt{\frac{5}{2}}\right)x + \left(\sqrt{\frac{5}{2} - 5}\right)y = 5$$

Find the equation of height drawn from vertex P: $y = -x$.

Orthocentre is the point of intersection of heights:

$$\left\{ \begin{array}{l} \left(5 + \sqrt{\frac{5}{2}}\right)x + \left(\sqrt{\frac{5}{2} - 5}\right)y = 5 \\ y = -x \end{array} \right.$$

Orthocentre is $(\alpha; \beta) = \left(\frac{1}{2}, -\frac{1}{2}\right)$. Then $|\alpha - 1| + |\beta - 1| = \left|\frac{1}{2} - 1\right| + \left| -\frac{1}{2} - 1 \right| = \frac{1}{2} + \frac{3}{2} = 2$

Answer: A) 2.

2. $\tan^4\frac{\pi}{16} +\cot^4\frac{\pi}{16} +\tan^4\frac{2\pi}{16} +\cot^4\frac{2\pi}{16} +\tan^4\frac{3\pi}{16} +\cot^4\frac{3\pi}{16}$ is equal to

(A) 645 (B) 646 (C) 848 (D) 678

Solution

\begin{array}{l} \tan^ {4} \frac {\pi}{1 6} + \cot^ {4} \frac {\pi}{1 6} + \tan^ {4} \frac {2 \pi}{1 6} + \cot^ {4} \frac {2 \pi}{1 6} + \tan^ {4} \frac {3 \pi}{1 6} + \cot^ {4} \frac {3 \pi}{1 6} = \\ = \frac {\sin^ {4} \frac {\pi}{1 6}}{\cos^ {4} \frac {\pi}{1 6}} + \frac {\cos^ {4} \frac {\pi}{1 6}}{\sin^ {4} \frac {\pi}{1 6}} + \frac {\sin^ {4} \frac {\pi}{8}}{\cos^ {4} \frac {\pi}{8}} + \frac {\cos^ {4} \frac {\pi}{8}}{\sin^ {4} \frac {\pi}{8}} + \frac {\sin^ {4} \frac {3 \pi}{1 6}}{\cos^ {4} \frac {3 \pi}{1 6}} + \frac {\cos^ {4} \frac {3 \pi}{1 6}}{\sin^ {4} \frac {3 \pi}{1 6}} = \\ = \frac {\left(\frac {1 - \cos \frac {\pi}{8}}{2}\right) ^ {2}}{\left(\frac {1 + \cos \frac {\pi}{8}}{2}\right) ^ {2}} + \frac {\left(\frac {1 + \cos \frac {\pi}{4}}{2}\right) ^ {2}}{\left(\frac {1 - \cos \frac {\pi}{4}}{2}\right) ^ {2}} + \frac {\left(\frac {1 + \cos \frac {\pi}{4}}{2}\right) ^ {2}}{\left(\frac {1 + \cos \frac {\pi}{4}}{2}\right) ^ {2}} + \frac {\left(\frac {1 + \cos \frac {\pi}{4}}{2}\right) + \left(\frac {1 - \cos \frac {\pi}{4}}{2}\right)}{2} + \\ + \frac {\left(\frac {1 - \cos \frac {3 \pi}{8}}{2}\right) ^ {2}}{\left(\frac {1 + \cos \frac {3 \pi}{8}}{2}\right) ^ {2}} + \frac {\left(\frac {1 + \cos \frac {3 \pi}{8}}{2}\right) ^ {2}}{\left(\frac {1 + \cos \frac {3 \pi}{8}}{2}\right) ^ {2}} = \frac {\left(\frac {1 - \sqrt {\frac {1 + \cos \frac {\pi}{4}}{2}}{2}\right) ^ {2}}{\left(\frac {1 + \sqrt {\frac {1 + \cos \frac {\pi}{4}}{2}}{2}\right) ^ {2}} + \\ \end{array} \left(\frac{1 + \sqrt{\frac{1 + \cos\frac{\pi}{4}}{2}}}{2}\right)^2 + \left(\frac{1 - \frac{\sqrt{2}}{2}}{2}\right)^2 + \left(\frac{1 + \frac{\sqrt{2}}{2}}{2}\right)^2 + \left(\frac{1 - \sqrt{\frac{1 + \cos\frac{3\pi}{4}}{2}}}{2}\right)^2 + \left(\frac{1 - \sqrt{\frac{1 + \cos\frac{\pi}{4}}{2}}}{2}\right)^2 + \left(\frac{1 + \sqrt{\frac{1 + \cos\frac{3\pi}{4}}{2}}}{2}\right)^2 + \left(\frac{1 + \sqrt{\frac{1 + \cos\frac{3\pi}{4}}{2}}}{2}\right)^2\right) + \left(\frac{1 + \sqrt{\frac{1 + \cos\frac{3\pi}{4}}{2}}}{2}\right)^2 \left(\frac{1 + \sqrt{\frac{1 + \cos\frac{3\pi}{4}}{2}}}{2}\right)^2 = \left(\frac{1 - \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}}{2}\right)^2 + \left(\frac{1 - \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}}{2}\right)^2 + \left(\frac{1 - \sqrt{\frac{1 + \sqrt{2}}{2}}{2}}\right)^2 + \left(\frac{1 - \sqrt{\frac{1 + \sqrt{2}}{2}}{2}}\right)^2 + \left(\frac{1 - \sqrt{\frac{1 + \sqrt{2}}{2}}{2}}\right)^2 + \left(\frac{1 - \sqrt{\frac{1 + \sqrt{2}}{2}}{2}}\right)^2 + \left(\frac{1 - \sqrt{\frac{1 + \sqrt{2}}{2}}{2}}\right)^2 + \left(\frac{1 - \sqrt{\frac{1 + \sqrt{2}}{2}}{2}}\right)^2 + \left(\frac{1 - \sqrt{\frac{1 + \sqrt{2}}{2}}{2}}\right)^2 + \left(\frac{1 - \sqrt{\frac{1 + \sqrt{2}}{2}}{2}}\right)^2 + \left(\frac{1 - \sqrt{\frac{1 + \sqrt{2}}{2}}{2}\right)^2 + \left(\frac{1 - \sqrt{\frac{1 + \sqrt{2}}{2}}{2}\right)^2 +

\begin{array}{l}

\left(\frac{1 - \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}}}{2}\right)^2 + \left(\frac{1 + \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}}}{2}\right)^2 \\

+ \left(\frac{1 + \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}}}{2}\right)^2 + \left(\frac{1 - \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}}}{2}\right)^2 = \frac{\left(2 - \sqrt{2 + \sqrt{2}}\right)^2}{(2 + \sqrt{2 + \sqrt{2}})^2} + \frac{\left(2 + \sqrt{2 + \sqrt{2}}\right)^2}{(2 - \sqrt{2 + \sqrt{2}})^2} + \\

+ \frac{(2 - \sqrt{2})^2}{(2 + \sqrt{2})^2} + \frac{(2 + \sqrt{2})^2}{(2 - \sqrt{2})^2} + \frac{(2 + \sqrt{2 - \sqrt{2}})}{(2 - \sqrt{2 - \sqrt{2}})^2} + \frac{(2 - \sqrt{2 - \sqrt{2}})^2}{(2 + \sqrt{2 - \sqrt{2}})^2} = 322 + 224\sqrt{2} + 322 - \\

-322 + 224\sqrt{2} + 34 = 678

\end{array}

Answer: D) 678. 3. If $f(x) = \tan x + \tan^2 x \cdot \tan 2x$ and $g(n) = \sum_{m=0}^{\infty} f\left(2^m\right)$, then $g(2013) - \tan\left(2^{2014}\right)$ is equal to (A) 0 (B) $\tan 1$ (C) $-\tan 1$ (D) None of these Solution

\sum_{n=0}^{2013} \tan(2^n) = \tan 1 + \tan 2 + \tan 4 + \dots + \tan 2013;

\sum_{n=0}^{2013} \tan^2(2^n) \tan(2 \cdot 2^n) = \tan^2 1 \tan 2 + \tan^2 2 \tan 4 + \tan^2 4 \tan 8 + \dots + \tan^2 2^{2013} \tan 2^{2014},

\sum_{n=0}^{2013} \tan(2^n) + \sum_{n=0}^{2013} \tan^2(2^n) \tan(2 \cdot 2^n) =

= \tan 1(1 + \tan 1 \tan 2) + \tan 2(1 + \tan 2 \tan 4) + \tan 4(1 + \tan 4 \tan 8) + \dots

+ \tan 2^{2013}(1 + \tan 2^{2013} \tan 2^{2014}) = \tan(2 - 1)(1 + \tan 1 \tan 2) + \tan(4 - 2)(1 + \tan 2 \tan 4) +

+ \tan(8 - 4)(1 + \tan 4 \tan 8) + \dots + \tan(2^{2014} - 2^{2013})(1 + \tan 2^{2013} \tan 2^{2014})

$$Using$$

\tan (\alpha - \beta) (1 + \tan \alpha \tan \beta) = \tan \alpha - \tan \beta

$$we obtain$$

\begin{array}{l}

\sum_{n=0}^{2013} \tan (2^n) + \sum_{n=0}^{2013} \tan^2 (2^n) \tan (2 \cdot 2^n) = \\

= \tan 2 - \tan 1 + \tan 4 - \tan 2 + \dots + \tan 2^{2014} - \tan 2^{2013} = - \tan 1 + \tan 2^{2014}

\end{array}

After subtracting $\tan(2^{2014})$ we obtain $-\tan 1$. Answer: C) $-\tan 1$. 4. The number of real roots of $\sqrt{5x^2 - 6x - 6} - \sqrt{5x^2 - 6x - 7} = 1$ is / are (A) 2 (B) 4 (C) 8 (D) None of these Solution

\sqrt{5x^2 - 6x - 6} - \sqrt{5x^2 - 6x - 7} = 1;

\sqrt{5x^2 - 6x - 6} = 1 + \sqrt{5x^2 - 6x - 7};

5x^2 - 6x - 6 = 1 + 2\sqrt{5x^2 - 6x - 7} + 5x^2 - 6x - 7;

\sqrt{5x^2 - 6x - 7} = 0;

5x^2 - 6x - 7 = 0;

x = \frac{3 + 2\sqrt{11}}{5}; \quad x = \frac{3 - 2\sqrt{11}}{5}.

Answer: A) 2. 5. Let $A$ be an average of numbers $2\sin 2{}^{\circ}$, $4\sin 4{}^{\circ}$, $6\sin 6{}^{\circ}, \ldots$, $180\sin 180{}^{\circ}$, then $A$ is equal to (A) $\cot 1{}^{\circ}$ (B) $\tan 1{}^{\circ}$ (C) $\sin 1{}^{\circ}$ (D) $\cos 1{}^{\circ}$ Solution

\sin 180{}^{\circ} = 0, \quad \sin (\alpha) = \sin (180{}^{\circ} - \alpha), \quad \sin 90{}^{\circ} = 1.

\begin{array}{l}

x = 2 \sin 2{}^{\circ} + 4 \sin 4{}^{\circ} + \dots + 90 \sin 90{}^{\circ} + \dots + 178 \sin 178{}^{\circ} = \\

= (2 + 178) \sin 2{}^{\circ} + (4 + 176) \sin 4{}^{\circ} + \dots = 180 (\sin 2{}^{\circ} + \sin 4{}^{\circ} + \dots + \sin 88{}^{\circ}) + 90 \sin 90{}^{\circ}

\end{array}

$$Then$$

\bar{x} = \frac{x}{90} = 2 \sin 2{}^{\circ} + 2 \sin 4{}^{\circ} + \dots + 2 \sin 88{}^{\circ} + 1;

\bar{x} \sin 1{}^{\circ} = 2 \sin 2{}^{\circ} \sin 1{}^{\circ} + 2 \sin 4{}^{\circ} \sin 1{}^{\circ} + \dots + 2 \sin 88{}^{\circ} \sin 1{}^{\circ} + \sin 1{}^{\circ};

$$Now$$

2 \sin 2{}^{\circ} \sin 1{}^{\circ} = \cos 1{}^{\circ} - \cos 3{}^{\circ};

2 \sin 4{}^{\circ} \sin 1{}^{\circ} = \cos 3{}^{\circ} - \cos 5{}^{\circ};

2 \sin 88{}^{\circ} \sin 1{}^{\circ} = \cos 87{}^{\circ} - \cos 89{}^{\circ}.

$$Hence$$

\bar{x} \sin 1{}^{\circ} = \cos 1{}^{\circ} - \cos 89{}^{\circ} + \sin 1{}^{\circ} = \cos 1{}^{\circ}.

$$Thus$$

\bar{x} = \cot 1{}^{\circ}.

$$

Answer: A) $\cot 1{}^{\circ}$.

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