heureka

Find the indicated power using De Moivre's Theorem. (Express your fully simplified answer in the form
\(a + bi\).)

\(\left(1-i\right)^{18}\)

\(\begin{array}{|rcll|} \hline 1-i &=& r\Big(\cos(x) -+ i \sin(x) \Big) \\ && \boxed{ r = \sqrt{1^2+(-1)^2} \\ r=\sqrt{2} } \\ && \boxed{ x = \arctan\left(\dfrac{-1}{1}\right) \\x = \arctan(-1)\\x = -45^\circ } \\ 1-i &=& \sqrt{2}\Big(\cos(-45^\circ) + i \sin(-45^\circ) \Big) \\ 1-i &=& \sqrt{2}\Big(\cos(45^\circ) - i \sin(45^\circ) \Big) \\ \left(1-i\right)^{18}&=& \left(\sqrt{2}\right)^{18}\Big(\cos(18*45^\circ) - i \sin(18*45^\circ) \Big) \quad | \quad \text{De Moivre's Theorem} \\ \left(1-i\right)^{18}&=& 2^{\frac{18}{2}}\Big(\cos(810^\circ) - i \sin(810^\circ) \Big) \\ \left(1-i\right)^{18}&=& 2^{9}\Big(\cos(90^\circ) - i \sin(90^\circ) \Big) \\ \left(1-i\right)^{18}&=& 512\Big(0 - i*1 \Big) \\ \mathbf{\left(1-i\right)^{18}} &=& \mathbf{-512i} \\ \hline \end{array}\)

Let A = (4, -1), B = (6, 2), and C = (-1, 2). There exists a point X and a constant k such that for any point P, \(PA^2 + PB^2 + PC^2 = 3PX^2 + k\). Find the constant \(k\).

Find the 3 smallest positive x-intercepts of the graph of \(y = \cos(12x) + \cos(14x)\) and list them in increasing order.

Formula: \(\boxed{\cos(x) + \cos(y) = 2\cos\left(\dfrac{x+y}{2}\right)\cos\left(\dfrac{x-y}{2}\right) }\)

\(\begin{array}{|rcll|} \hline && \mathbf{\cos(12x) + \cos(14x)} \\ &=& \cos(14x) + \cos(12x) \\ &=& 2\cos\left(\dfrac{14x+12x}{2}\right)\cos\left(\dfrac{14x-12x}{2}\right) \\ &=& 2\cos\left(\dfrac{26x}{2}\right)\cos\left(\dfrac{2x}{2}\right) \\ &=& \mathbf{2\cos(13x)\cos(x)} \\ \hline \end{array}\)

 x-intercepts of the graph

\(\begin{array}{|rcll|} \hline \mathbf{2\cos(13x)\cos(x)} &=& 0 \quad | \quad : 2 \\ \cos(13x)\cos(x) &=& 0 \\ \hline \mathbf{\cos(x)} &=& \mathbf{0} \\ x &=& \pm \arccos(0) +2k\pi \qquad k\in \mathbb{Z} \\ x &=& \pm \dfrac{\pi}{2} +2k\pi \\\\ x &=& \mathbf{+} \dfrac{\pi}{2}+2k\pi \qquad k = 0 \\ x &=& + \dfrac{\pi}{2} \\ \mathbf{x} &=& \mathbf{1.57079632679}\ \text{rad} \\\\ x &=& \mathbf{-} \dfrac{\pi}{2} +2k\pi \qquad k = 1 \\ x &=& - \dfrac{\pi}{2} +2 \pi \\ \mathbf{x} &=& \mathbf{4.71238898038}\ \text{rad} \\ \hline \mathbf{\cos(13x)} &=& \mathbf{0} \\ 13x &=& \pm \arccos(0) +2k\pi \qquad k\in \mathbb{Z} \\ 13x &=& \pm \dfrac{\pi}{2} +2k\pi \\\\ 13x &=& \mathbf{+}\dfrac{\pi}{2} +2k\pi \\ x &=& \dfrac{\pi}{2*13}+\dfrac{2\pi k}{13} \qquad k = 0 \\ x &=& \dfrac{\pi}{26} \\ \mathbf{x} &=& \mathbf{0.12083048668}\ \text{rad} \\\\ x &=& \mathbf{+}\dfrac{\pi}{26}+\dfrac{2\pi k}{13} \qquad k = 1 \\ x &=& \dfrac{\pi}{26}+\dfrac{2\pi}{13} \\ x &=& \dfrac{5\pi}{26} \\ \mathbf{x} &=& \mathbf{0.60415243338}\ \text{rad} \\\\ 13x &=& \mathbf{-}\dfrac{\pi}{2} +2k\pi \\ x &=& -\dfrac{\pi}{2*13}+\dfrac{2\pi k}{13} \qquad k = 1 \\ x &=& -\dfrac{\pi}{26}+\dfrac{2\pi}{13} \\ x &=& \dfrac{3\pi}{26} \\ \mathbf{x} &=& \mathbf{0.36249146003}\ \text{rad} \\ \hline \end{array}\)

The 3 smallest positive x-intercepts of the graph:
\(\mathbf{\dfrac{\pi}{26} },\ \mathbf{\dfrac{3\pi}{26} },\ \mathbf{\dfrac{5\pi}{26} } \)

A mathematician works for \(t\) hours per day and solves \(p\) problems per hour, where \(t\) and \(p\) are positive integers and \(1 < p < 20\) .
One day, the mathematician drinks some coffee and discovers that he can now solve \(3p+7\) problems per hour.
In fact, he only works for \(t-4\) hours that day, but he still solves twice as many problems as he would in a normal day.
How many problems does he solve the day he drinks coffee?

Formula: \((3p+7)(t-4) = 2tp\)

\(\begin{array}{|rcll|} \hline (3p+7)(t-4) &=& 2tp \\ 3pt-12p+7t-28 &=& 2tp \\ pt-12p+7t-28 &=& 0 \\ t(p+7)-12p-28 &=& 0 \\ t(p+7)&=& 12p+28 \\ t(p+7)&=& 4(3p+7) \\ t &=& 4\left(\dfrac{3p+7}{p+7}\right) \\ t &=& 4\left(\dfrac{3p+21-14}{p+7}\right) \\ t &=& 4\left(\dfrac{3(p+7)-14}{p+7}\right) \\ \mathbf{t} &=& \mathbf{4\left(3-\dfrac{14}{p+7}\right)} \\ \hline \end{array} \)

\(t\) and \(p\) are positive integers

\(\begin{array}{|rcll|} \hline \dfrac{14}{p+7} &=& 0 \qquad t > 0 \checkmark \\ \dfrac{14}{p+7} &=& 1 \qquad t > 0 \checkmark \\ \dfrac{14}{p+7} &=& 2 \qquad t > 0 \checkmark \\ \dfrac{14}{p+7} &=& 3 \qquad t = 0 \\ \dfrac{14}{p+7} &>& 3 \qquad t < 0 \\ \hline \end{array} \)

\(\mathbf{p=\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{14}{p+7}} &=& \mathbf{0} \\ 14 &=& 0(p+7) \\ 14 &=& 0 \qquad | \text{ does not yield any value of p.} \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \mathbf{\dfrac{14}{p+7}} &=& \mathbf{1} \\ 14 &=& 1(p+7) \\ 14 &=& p+7 \\ \mathbf{p} &=& \mathbf{7}\ \checkmark \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \mathbf{\dfrac{14}{p+7}} &=& \mathbf{2} \\ 14 &=& 2(p+7) \\ 14 &=& 2p+14 \\ 0 &=& 2p \\ p &=& 0\qquad 1 < p < 20,\ \text{ no solution, p > 1!} \\ \hline \end{array}\)

\(\mathbf{t=\ ?}\)

\(\begin{array}{|rcll|} \hline t &=& 4\left(\dfrac{3p+7}{p+7}\right) \quad | \quad \mathbf{p=7} \\ t &=& 4\left(\dfrac{3*7+7}{7+7}\right) \\ t &=& 4\left(\dfrac{4*7}{2*7}\right) \\ t &=& 4\left(\dfrac{4 }{2 }\right) \\ t &=& 4*2 \\ \mathbf{t} &=& \mathbf{8} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 2tp &=& 2*8*7 \\ \mathbf{2tp} &=& \mathbf{112} \\ \hline \end{array}\)

The mathematician solved 112 problems the day he drank coffee.

A mathematician works for \(t\) hours per day and solves \(p\) problems per hour, where \(t\) and \(p\) are positive integers and \(1< p < 20\) .
One day, the mathematician drinks some coffee and discovers that he can now solve \(3p+7\) problems per hour.
In fact, he only works for \(t-4\) hours that day, but he still solves twice as many problems as he would in a normal day.
How many problems does he solve the day he drinks coffee?

James’s four-year-old brother is trying to arrange black and yellow Legos in rows.
He has 56 yellow Legos and 120 black Legos.
He wants to arrange them in rows so that there are either only yellow or only black Legos in each row,
and so that each row has the same number of Legos.
What is the least number of rows he can make?

\(\begin{array}{|rcll|} \hline gcd(56,120) &=& 8 \\ \dfrac{56}{8} &=& 7 \\ \dfrac{120}{8} &=& 15 \\ \hline \end{array}\)

The least number of rows he can have is 22;

7 rows of yellow and 15 rows of black.

pls help
\(\text{$\cos(\theta) = -\dfrac{\sqrt{2}}{3}$, where $\pi \leq \theta\leq \dfrac{3\pi}{2}$. [quadrant $III$] } \)
\(\text{$\tan(\beta) = \dfrac{4}{3}$, where $0 \leq \beta \leq \dfrac{\pi}{2}$. [quadrant $I$] }\)
What is the exact value of \(\sin(\theta+\beta) \)

\(\begin{array}{rcll} \cos(\theta_{III}) &=& -\dfrac{\sqrt{2}}{3} \\ \tan(\beta_{I}) &=& \dfrac{4}{3} \\ \end{array}\)

\(\begin{array}{|rcll|} \hline \sin(\theta_{III}) &=& - \sqrt{1-\cos^2(\theta_{III})} \quad | \quad \boxed{\sin(\theta_{III}) < 0\ !} \\ \sin(\theta_{III}) &=& - \sqrt{1- \left(-\dfrac{\sqrt{2}}{3} \right)^2 } \\ \sin(\theta_{III}) &=& - \sqrt{1- \dfrac{2}{9} } \\ \sin(\theta_{III}) &=& - \sqrt{\dfrac{9-2}{9} } \\ \sin(\theta_{III}) &=& - \sqrt{\dfrac{7}{9} } \\ \mathbf{\sin(\theta_{III})} &=& \mathbf{- \dfrac{\sqrt{7}}{3}} \\ \hline \end{array}\)

Formula:

\(\begin{array}{|rcll|} \hline \sin^2(\beta)+\cos^2(\beta) &=& 1 \quad | \quad : \cos^2(\beta) \\ \tan^2(\beta)+1 &=& \dfrac{1}{\cos^2(\beta)} \\ \cos^2(\beta) &=& \dfrac{1}{1+\tan^2(\beta)} \\ \mathbf{\cos(\beta)} &=& \mathbf{\pm\sqrt{ \dfrac{1}{1+\tan^2(\beta)} } } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \cos(\beta_{I}) &=& +\sqrt{ \dfrac{1}{1+\tan^2(\beta_{I})} } \quad | \quad \boxed{\cos(\beta_{I}) > 0\ !} \\ \cos(\beta_{I}) &=& \sqrt{ \dfrac{1}{1+ \left(\dfrac{4}{3}\right)^2} } \\ \cos(\beta_{I}) &=& \sqrt{ \dfrac{1}{1+ \dfrac{16}{9}} } \\ \cos(\beta_{I}) &=& \sqrt{ \dfrac{1}{\dfrac{9+16}{9}} } \\ \cos(\beta_{I}) &=& \sqrt{ \dfrac{1}{\dfrac{25}{9}} } \\ \cos(\beta_{I}) &=& \sqrt{ \dfrac{9}{25} } \\ \mathbf{\cos(\beta_{I})} &=& \mathbf{\dfrac{3}{5}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \sin(\beta_{I}) &=& \tan(\beta_{I})\cos(\beta_{I}) \\ \sin(\beta_{I}) &=& \left(\dfrac{4}{3}\right)\left(\dfrac{3}{5}\right) \\ \mathbf{\sin(\beta_{I})} &=& \mathbf{\dfrac{4}{5}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\sin(\theta_{III}+\beta_{I}) } &=& \mathbf{\sin(\theta_{III})\cos(\beta_{I})+\cos(\theta_{III})\sin(\beta_{I}) } \\ \\ \sin(\theta_{III}+\beta_{I}) &=& \left( - \dfrac{\sqrt{7}}{3} \right) \left( \dfrac{3}{5} \right) + \left( -\dfrac{\sqrt{2}}{3}\right) \left( \dfrac{4}{5}\right) \\ \sin(\theta_{III}+\beta_{I}) &=& -\dfrac{3\sqrt{7}}{15} -\dfrac{4\sqrt{2}}{15} \\ \mathbf{\sin(\theta_{III}+\beta_{I})} &=& \mathbf{-\dfrac{\sqrt{7}}{5} -\dfrac{4\sqrt{2}}{15}} \\ \hline \end{array}\)

Angle bisectors \(\overline{AX}\) and \(\overline{BY}\) of triangle \(ABC\) meet at point \(I\).
Find \(\angle C\), in degrees, if \(\angle AIB = 109^\circ\).

\(\begin{array}{|rcll|} \hline \text{In }\angle ABC \\ \hline 2x+2y +C &=& 180^\circ \\ 2x+2y &=& 180^\circ-C \\ 2(x+y) &=& 180^\circ-C \\ \mathbf{x+y} &=& \mathbf{\dfrac{180^\circ-C}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{In } IXCY \\ \hline 360^\circ &=& 109^\circ + \Big(180^\circ-(C+x)\Big) + C + \Big(180^\circ-(C+y)\Big) \\ 360^\circ &=& 109^\circ + 180^\circ-C-x + C + 180^\circ-C-y \\ 0 &=& 109^\circ -C-x + C -C-y \\ 0 &=& 109^\circ -C-x -y \\ C &=& 109^\circ -x -y \\ C &=& 109^\circ -(x+y) \quad | \quad \mathbf{x+y=\dfrac{180^\circ-C}{2}} \\ C &=& 109^\circ - \left(\dfrac{180^\circ-C}{2}\right) \quad | \quad * 2 \\ 2C &=& 218^\circ - (180^\circ-C) \\ 2C &=& 218^\circ - 180^\circ + C \quad | \quad -C \\ C &=& 218^\circ - 180^\circ \\ \mathbf{C} &=& \mathbf{38^\circ} \\ \hline \end{array}\)

Find sin2x, cos2x, and tan2x if cos(x) = 1/sqrt(10) and x terminates in quadrant IV.

\(\begin{array}{|rcll|} \hline \cos(x) &=& \dfrac{1}{\sqrt{10}} \\ \cos(x) &=& 0.31622776602 \\ x &=& \pm\arccos(0.31622776602) \\ x &=& \pm71.5650511771^\circ \\ x_1 &=& +71.5650511771^\circ \quad | \quad \text{quadrant $I$} \\ \mathbf{x_2} &=& \mathbf{-71.5650511771^\circ} \quad | \quad \text{quadrant $IV$} \\ \hline \end{array}\)

\(\text{$\mathbf{x}$ terminates in quadrant $IV$ and is $\mathbf{-71.5650511771^\circ}$}\)

\(\begin{array}{|rcll|} \hline \sin(2x) &=& \sin\Big(2*(-71.5650511771^\circ)\Big) \\ \sin(2x) &=& \sin (-143.130102354^\circ) \\ \mathbf{\sin(2x)} &=& \mathbf{-0.6} \\ \hline \end{array} \begin{array}{|rcll|} \hline \cos(2x) &=& \cos\Big(2*(-71.5650511771^\circ)\Big) \\ \cos(2x) &=& \cos (-143.130102354^\circ) \\ \mathbf{\cos(2x)} &=& \mathbf{-0.8} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \tan(2x) &=& \tan\Big(2*(-71.5650511771^\circ)\Big) \\ \tan(2x) &=& \tan (-143.130102354^\circ) \\ \mathbf{\tan(2x)} &=& \mathbf{0.75} \\ \hline \end{array}\)

Let \(x\) and \(y\) be positive real numbers.
Define \(a = 1 + \dfrac{x}{y}\) and  \(b = 1 + \dfrac{y}{x}\).
If \(a^2 + b^2 = 15\) compute \(a^3 + b^3\)

\(\begin{array}{|lrcll|} \hline (1) & a &=& 1 + \dfrac{x}{y} \\ & \mathbf{ \dfrac{x}{y} } &=& \mathbf{a-1} \\\\ (2) & b &=& 1 + \dfrac{y}{x} \\ & \dfrac{y}{x}&=& b-1 \\ &\mathbf{\dfrac{x}{y}} &=& \mathbf{\dfrac{1}{ b-1}} \\\\ \hline & \mathbf{\dfrac{x}{y} =} a-1 &=& \dfrac{1}{ b-1} \\ & a-1 &=& \dfrac{1}{ b-1} \\ & (a-1)(b-1) &=& 1 \\ & ab-(a+b)+1 &=& 1 \\ & \mathbf{ab} &=& \mathbf{a+b} \\ \hline & (ab)^2 &=& (a+b)^2 \\ & (ab)^2 &=& a^2+2ab+b^2 \\ & (ab)^2 &=& a^2+b^2+ 2ab \quad | \quad a^2 + b^2 = 15 \\ & (ab)^2 &=& 15+ 2ab \\ & \mathbf{(ab)^2 - 2ab -15} &=& \mathbf{ 0 } \\\\ & ab &=& \dfrac{2\pm\sqrt{4-4*(-15)}} {2} \\ & ab &=& \dfrac{2\pm\sqrt{4+4*15}} {2} \\ & ab &=& \dfrac{2\pm\sqrt{64}} {2} \\ & ab &=& \dfrac{2\pm 8} {2} \\ & ab &=& \dfrac{2 + 8} {2} \\ &&& \text{$a$ and $b$ be positive, while $x$ and $y$ be positive}, \\ &&& \text{so $ab$ be positive } \\ & \mathbf{ab} &=& \mathbf{5} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a^3+b^3 &=& (a+b)(a^2-ab+b^2) \\ a^3+b^3 &=& (a+b)(a^2+b^2-ab) \quad & | \quad a^2+b^2 = 15 \\ a^3+b^3 &=& (a+b)(15-ab) \quad & | \quad a+b=ab=5 \\ a^3+b^3 &=& 5(15-5) \\ a^3+b^3 &=& 5*10 \\ \mathbf{a^3+b^3} &=& \mathbf{50} \\ \hline \end{array}\)