heureka

arithmetic series

Evaluate:  \(\sum \limits_{n=1}^{50} (4n-1)\)
 

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{n=1}^{50} (4n-1)} \\\\ &=& \sum \limits_{n=1}^{50} (4n) - \sum \limits_{n=1}^{50} (1) \\\\ &=& \sum \limits_{n=1}^{50} (4n) - 50\times 1 \\\\ &=& \sum \limits_{n=1}^{50} (4n) - 50 \\\\ &=& 4\sum \limits_{n=1}^{50} (n) - 50 \\\\ &=& 4\left(1+2+3+\ldots + 49+50 \right) - 50 \\\\ &=& 4\left( \dfrac{(1+50)}{2}\times 50 \right) - 50 \\\\ &=& 2\times 51\times 50 - 50 \\\\ &=& 5100 - 50 \\\\ &=& \mathbf{5050} \\ \hline \end{array}\)

Let \(A\) and \(B\) be invertible matrices such that \(\mathbf{A} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \text{ and } \mathbf{B} \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}\).
Calculate \((\mathbf{A}\mathbf{B})^{-1} \begin{pmatrix} 1 \\ 2 \end{pmatrix}\).

\(\begin{array}{lcll} \text{Let }v_1 &=& \begin{pmatrix} 2 \\ -1 \end{pmatrix} \\\\ \text{Let }v_2 &=& \begin{pmatrix} 1 \\ 2 \end{pmatrix} \\\\ \text{Let }v_3 &=& \begin{pmatrix} 1 \\ 3 \end{pmatrix} \\\\ \text{Let }Av_1 &=& v_2 \\\\ \text{Let }Bv_3 &=& v_1 \\ \end{array}\)

Formula: \(\boxed{(AB)^{-1}=B^{-1}A^{-1}}\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ (\mathbf{A}\mathbf{B})^{-1} \begin{pmatrix} 1 \\ 2 \end{pmatrix} } \\ &=& (\mathbf{A}\mathbf{B})^{-1} v_2 \quad &| \quad \boxed{(AB)^{-1}=B^{-1}A^{-1}} \\ &=& B^{-1}A^{-1}v_2 \quad &| \quad v_2=Av_1 \\ &=& B^{-1}A^{-1} Av_1 \quad &| \quad A^{-1} A=I\quad \text{identity matrix} \\ &=& B^{-1}Iv_1 \quad &| \quad Iv_1=v_1 \\ &=& B^{-1}v_1 \quad &| \quad v_1=Bv_3 \\ &=& B^{-1}Bv_3 \quad &| \quad B^{-1} B=I\quad \text{identity matrix} \\ &=& Iv_3 \quad &| \quad Iv_3=v_3 \\ &=&\mathbf{ v_3 } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{ (\mathbf{A}\mathbf{B})^{-1} \dbinom12 = \dbinom13 } \\ \hline \end{array}\)

How do you find the quadratic equation if the roots are \(x = 3 + \sqrt{5}\) and \(x = 3 - \sqrt{5}\)?

\(\begin{array}{|rcll|} \hline && (x-r_1)(x-r_2) \\ &=& \Big(x-(3 - \sqrt{5})\Big)\Big(x-(3 + \sqrt{5})\Big) \\ &=& (x-3 + \sqrt{5})(x-3 - \sqrt{5}) \\ &=& \Big( (x-3) + \sqrt{5}\Big) \Big((x-3) - \sqrt{5}\Big) \\ &=& (x-3)^2 - (\sqrt{5})^2 \\ &=& x^2-6x+9-5 \\ &=& \mathbf{ x^2-6x+4 } \\ \hline \end{array}\)

Find constants \(A\) and \(B\) such that \(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{A}{x - 2} + \dfrac{B}{x + 1}\)
for all \(x\) such that \(x\neq -1\) and \(x\neq 2\).
Give your answer as the ordered pair \((A,\ B)\).

\(\text{Let $x^2 - x - 2 = (x-2)(x+1)$}\)

\(\begin{array}{|rcll|} \hline \dfrac{x + 7}{x^2 - x - 2} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad & | \quad \cdot (x-2)(x+1) \\\\ x + 7 &=& \dfrac{A(x-2)(x+1)}{x - 2} + \dfrac{B(x-2)(x+1)}{x + 1} \quad & | \quad x\ne2,\ x\ne -1 \\\\ x + 7 &=& A (x+1) + B(x-2) \\ \\ \hline x=-1: \quad -1 + 7 &=& A (-1+1) + B(-1-2) \\ \quad 6 &=& 0 -3B \\ \quad 6 &=& -3B \\ \quad -3B &=& 6 \quad | \quad :(-3) \\ \quad B &=& \frac{6}{-3} \\ \quad \mathbf{B} & \mathbf{=} & \mathbf{-2} \\ \hline x=2: \quad 2 + 7 &=& A (2+1) + B(2-2) \\ \quad 9 &=& 3A +0 \\ \quad 9 &=& 3A \\ \quad 3A &=& 9 \quad | \quad :3 \\ \quad A &=& \frac{9}{3} \\ \quad \mathbf{A} & \mathbf{=} & \mathbf{3} \\ \hline \end{array}\)

\(\mathbf{(A,\ B) = (3,\ -2)}\)

Let \(A\) be a matrix such that \(\mathbf{A}^{-1} = \begin{pmatrix} 1 & 1 \\ 2 & x \end{pmatrix}\)
for some value of \(x\) .
What is the vector that  maps to \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\)?

I assume:

\(\begin{array}{|rcll|} \hline Av &=& \dbinom10 \quad & | \quad \times A^{-1}\\ A^{-1}Av &=& A^{-1}\dbinom10 \quad & | \quad A^{-1}A = I \\ Iv &=& A^{-1}\dbinom10 \quad & | \quad Iv = v \\ v &=& A^{-1}\dbinom10 \\\\ v &=& \begin{pmatrix} 1 & 1 \\ 2 & x \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} \\\\ v &=& \begin{pmatrix} 1*1+1*0 \\ 2*1+x*0 \end{pmatrix} \\\\ \mathbf{v} &=& \mathbf{ \dbinom12 } \\ \hline \end{array}\)

The quantity \(\tan 7.5^\circ\) can be expressed in the form \(\tan 7.5^\circ = \sqrt{a} - \sqrt{b} + \sqrt{c} - d\),
where \(a \ge b \ge c \ge d\) are positive integers.
Find \(a + b + c + d\).

Formula: \(\boxed {\cos(2\varphi) = \cos^2(\varphi)-\sin^2(\varphi) \\ \Rightarrow \quad \sin^2(\varphi)=\dfrac{1-\cos(2\varphi)}{2},\quad \cos^2(\varphi)=\dfrac{1+\cos(2\varphi)}{2} }\)

\(\text{Let $\cos(30^\circ)=\dfrac{\sqrt{3}}{2} $}\)

\(\begin{array}{|rcll|} \hline \mathbf{\sin^2(15^\circ)} &=& \mathbf{\dfrac{1-\cos(30^\circ)}{2}} \\ \sin^2(15^\circ) &=& \dfrac{1-\dfrac{\sqrt{3}}{2}}{2} \\ \mathbf{ \sin^2(15^\circ) } &=& \mathbf{\dfrac{1}{2}-\dfrac{\sqrt{3}}{4}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\cos^2(15^\circ)} &=& \mathbf{\dfrac{1+\cos(30^\circ)}{2}} \\ \cos^2(15^\circ) &=& \dfrac{1+\dfrac{\sqrt{3}}{2}}{2} \\ \mathbf{ \cos^2(15^\circ) } &=& \mathbf{\dfrac{1}{2}+\dfrac{\sqrt{3}}{4}} \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \mathbf{\tan^2(15^\circ)} &=& \mathbf{\dfrac{\sin^2(15^\circ)} {\cos^2(15^\circ)}} \\\\ \tan^2(15^\circ) &=& \dfrac{\left( \dfrac{1}{2}-\dfrac{\sqrt{3}}{4}\right) } { \left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{4} \right)} \\\\ \tan^2(15^\circ) &=& \dfrac{\left( \dfrac{1}{2}-\dfrac{\sqrt{3}}{4}\right) } { \left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{4} \right)}\times \dfrac{\left( \dfrac{1}{2}-\dfrac{\sqrt{3}}{4}\right) } { \left( \dfrac{1}{2}-\dfrac{\sqrt{3}}{4} \right)} \\\\ \tan^2(15^\circ) &=& \dfrac{\left( \dfrac{1}{4}-\dfrac{\sqrt{3}}{4}+\dfrac{3}{16} \right) } { \left( \dfrac{1}{4}-\dfrac{3}{16} \right)} \\\\ \tan^2(15^\circ) &=& 7 - 4\sqrt{3} \\ \hline \tan (15^\circ) &=& \sqrt{ 7 - 4\sqrt{3}} \\ \tan (15^\circ) &=& \sqrt{ 7 - \sqrt{16*3}} \\ \mathbf{\tan (15^\circ)} &=& \mathbf{\sqrt{ 7 - \sqrt{48}} } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{\tan (15^\circ)} &=& \mathbf{\sqrt{ 7 - \sqrt{48}} } \\\\ \sqrt{ 7 - \sqrt{48}} &=& \sqrt{u}-\sqrt{v} \\ \ 7 - \sqrt{48} &=& \left(\sqrt{u}-\sqrt{v}\right)^2 \\ \ 7 - \sqrt{48} &=& u+2\sqrt{uv}+v \\ \ 7 - \sqrt{48} &=& u+v+\sqrt{4uv} \quad \text{compare}\\ && \begin{array}{|rcll|} \hline u+v &=&7 \quad \text{ or }\quad v = 7-u \\ 4uv&=&48 \\ 4u(7-u) &=& 48 \\ u(7-u) &=& 12 \\ u^2 -7u+12 &=& 0 \\\\ u &=& \dfrac{7\pm \sqrt{49-4*12} }{2} \\ u &=& \dfrac{7\pm 1 }{2} \\ \mathbf{u} &=& \mathbf{4} \\\\ v &=& 7-u \\ v &=& 7-4 \\ \mathbf{v} &=& \mathbf{3} \\ \hline \end{array} \\ \mathbf{\tan (15^\circ)} &=& \mathbf{\sqrt{u}-\sqrt{v} } \\ \tan (15^\circ) &=& \sqrt{4}-\sqrt{3} \\ \mathbf{\tan (15^\circ)} &=& \mathbf{2-\sqrt{3}} \\ \hline \dfrac{1}{\tan (15^\circ)} &=& \left(\dfrac{1}{2-\sqrt{3}} \right) \times \left(\dfrac{2+\sqrt{3}}{2+\sqrt{3}} \right) \\\\ \dfrac{1}{\tan (15^\circ)} &=& \dfrac{2+\sqrt{3}}{4-3} \\\\ \mathbf{\dfrac{1}{\tan (15^\circ)}} &=& \mathbf{2+\sqrt{3}} \\ \hline \end{array}\)

Formula: \(\boxed {\tan(2\varphi) = \dfrac{2\tan(\varphi)}{1-\tan^2(\varphi)} }\)

\(\begin{array}{|rcll|} \hline \mathbf{\tan(2\varphi)} &=& \mathbf{\dfrac{2\tan(\varphi)}{1-\tan^2(\varphi)}} \\\\ \tan(15^\circ) &=& \dfrac{2\tan(7.5^\circ)}{1-\tan^2(7.5^\circ)} \\\\ \tan(15^\circ)\left(1-\tan^2(7.5^\circ)\right) &=& 2\tan(7.5^\circ) \\ \tan(15^\circ) -\tan(15^\circ)\tan^2(7.5^\circ) &=& 2\tan(7.5^\circ) \\ \tan(15^\circ)\tan^2(7.5^\circ)+ 2\tan(7.5^\circ) -\tan(15^\circ) &=& 0 \quad | \quad : \tan(15^\circ) \\ \tan^2(7.5^\circ)+ 2\times\dfrac{1}{\tan(15^\circ)} \tan(7.5^\circ) - 1 &=& 0 \quad | \quad \mathbf{\dfrac{1}{\tan (15^\circ)}=2+\sqrt{3}} \\ \mathbf{\tan^2(7.5^\circ)+ 2(2+\sqrt{3}) \tan(7.5^\circ) - 1} &=& \mathbf{0} \\\\ \tan(7.5^\circ) &=& \dfrac{-2(2+\sqrt{3}) \pm \sqrt{4(2+\sqrt{3})^2-4(-1)} }{2} \\\\ \tan(7.5^\circ) &=& \dfrac{-2(2+\sqrt{3}) \pm \sqrt{4(2+\sqrt{3})^2+4} }{2} \\\\ \tan(7.5^\circ) &=& \dfrac{-2(2+\sqrt{3}) \pm \sqrt{4\left( (2+\sqrt{3})^2+1\right)} }{2} \\\\ \tan(7.5^\circ) &=& \dfrac{-2(2+\sqrt{3}) \pm 2\sqrt{(2+\sqrt{3})^2+1} }{2} \\\\ \tan(7.5^\circ) &=& -2-\sqrt{3} \pm \sqrt{(2+\sqrt{3})^2+1} \\\\ \tan(7.5^\circ) &=& -2-\sqrt{3} \pm \sqrt{8+4\sqrt{3}} \\ \tan(7.5^\circ) &=& -2-\sqrt{3} \pm \sqrt{8+\sqrt{16*3}} \\ \mathbf{\tan(7.5^\circ)} &=& \mathbf{-2-\sqrt{3} + \sqrt{8+\sqrt{48}}} \quad | \quad \tan(7.5^\circ) > 0\ ! \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{ \sqrt{8+\sqrt{48} } } &=& \mathbf{\sqrt{u}+\sqrt{v}} \\ \ 8+\sqrt{48} &=& \left(\sqrt{u}+\sqrt{v}\right)^2 \\ \ 8+\sqrt{48} &=& u+2\sqrt{uv}+v \\ \ 8+\sqrt{48} &=& u+v+\sqrt{4uv} \quad \text{compare}\\ && \begin{array}{|rcll|} \hline u+v &=&8 \quad \text{ or }\quad v = 8-u \\ 4uv&=&48 \\ 4u(8-u) &=& 48 \\ u(8-u) &=& 12 \\ u^2 -8u+12 &=& 0 \\\\ u &=& \dfrac{8\pm \sqrt{64-4*12} }{2} \\ u &=& \dfrac{8\pm 4 }{2} \\ \mathbf{u} &=& \mathbf{6} \\\\ v &=& 8-u \\ v &=& 8-6 \\ \mathbf{v} &=& \mathbf{2} \\ \hline \end{array} \\ \mathbf{ \sqrt{8+\sqrt{48}} }&=& \mathbf{ \sqrt{u}+\sqrt{v} } \\ \mathbf{ \sqrt{8+\sqrt{48}} } &=& \mathbf{ \sqrt{6}+\sqrt{2} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\tan(7.5^\circ)} &=& \mathbf{-2-\sqrt{3} + \sqrt{8+\sqrt{48}}} \quad & | \quad \mathbf{ \sqrt{8+\sqrt{48}}=\sqrt{6}+\sqrt{2} } \\ \tan(7.5^\circ) &=& -2-\sqrt{3} + \sqrt{6}+\sqrt{2} \\ \mathbf{ \tan(7.5^\circ) } &=& \mathbf{\sqrt{6}-\sqrt{3}+\sqrt{2}-2} \\ \hline a+b+c+d &=& 6+3+2+2 \\ \mathbf{a+b+c+d} &=& \mathbf{13} \\ \hline \end{array}\)

Solve the linear Diophantine equation 11x - 13y = 1 using Euclid's algorithm.

\(\begin{array}{|rcll|} \hline 11x-13y=1 \\ \mathbf{11x+(-13)y=1} \\ \text{Euclid's Algorithm} &&&\text{remainder}\\ \hline -13 &=& -1\times11-2 & -2 = -13+11 \\ 11 &=& -5\times(-2)+1 & 1 =11+5\times(-2) \\ -2 &=& -2\times 1+0 \qquad \text{stop} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{backwards substitution} \\ \hline 1 &=& 11+5\times(-2) \quad & | \quad -2=-13+11 \\ 1 &=& 11+5\times(-13+11) \\ 1 &=& 11+5\times(-13) +5\times+11 \\ 1 &=& 6\times 11+5\times(-13) \\ \hline x=6 && y=5 \\ &\text{check}& \\ 11\times 6 +5\times(-13) &=& 1\\ 66-65 &=& 1 \ \checkmark \\ \hline \end{array} \)

P is a point outside of equilateral triangle ABC. PA=3, PB=4, PC=5. Find AB. 

cos rule:

\(\begin{array}{|rcll|} \hline 4^2 &=& 3^2+x^2-2*3*x\cos(y) \\ 16 &=& 9+x^2-6x\cos(y) \\ 6x\cos(y) &=& -16 + 9+x^2 \\ 6x\cos(y) &=& x^2-7 \\ \mathbf{\cos(y)} &=& \mathbf{\dfrac{x^2-7}{6x}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \sin^2(y) &=& 1-\cos^2(y) \\ \sin^2(y) &=& 1-\dfrac{(x^2-7)^2}{36x^2} \\ \sin^2(y) &=& \dfrac{36x^2-(x^2-7)^2}{36x^2} \\ \sin(y) &=& \dfrac{ \sqrt{36x^2-(x^2-7)^2} } {6x} \\ \sin(y) &=& \dfrac{ \sqrt{36x^2-x^4+14x^2-49} } {6x} \\ \mathbf{\sin(y)} &=& \mathbf{\dfrac{1}{6x} \sqrt{50x^2-x^4-49 } } \\ \hline \end{array} \)

cos rule:

\(\begin{array}{|rcll|} \hline 5^2&=& 3^2+x^2-2*3x\cos(60^\circ+y) \\ 25&=& 9+x^2-6x\cos(60^\circ+y) \\ 6x\cos(60^\circ+y) &=& x^2-16 \\ \mathbf{\cos(60^\circ+y)} &=& \mathbf{\dfrac{x^2-16} {6x}} \\\\ \cos(60^\circ)\cos(y)-\sin(60^\circ)\sin(y) &=& \dfrac{x^2-16} {6x} \\ \dfrac{1}{2}\cos(y)-\dfrac{\sqrt{3}}{2}\sin(y) &=& \dfrac{x^2-16} {6x} \quad | \quad \times 2 \\ \cos(y)- \sqrt{3}\sin(y) &=& \dfrac{2x^2-32} {6x} \\ \dfrac{x^2-7}{6x}- \sqrt{3}\dfrac{1}{6x} \sqrt{50x^2-x^4-49 } &=& \dfrac{2x^2-32} {6x} \quad | \quad \times 6x \\ x^2-7 - \sqrt{3} \sqrt{50x^2-x^4-49 } &=& 2x^2-32 \\ -\sqrt{3}\sqrt{50x^2-x^4-49 } &=& x^2-25 \qquad \text{square both sides} \\ 3(50x^2-x^4-49) &=& (x^2-25)^2 \\ 150x^2-3x^4-147 &=& x^4-50x^2+625 \\ \ldots \\ 4x^4-200x^2+772 &=& 0 \quad | \quad : 4 \\ \mathbf{x^4-50x^2+193} &=& \mathbf{0} \\\\ x^2 &=& \dfrac{50\pm \sqrt{50^2-4*193}} {2} \\ x^2 &=& \dfrac{50\pm \sqrt{4*432}} {2} \\ x^2 &=& \dfrac{50\pm 2\sqrt{432}} {2} \\ x^2 &=& 25 \pm \sqrt{432} \\ x &=& \sqrt{ 25 \pm \sqrt{432} } \\ x &=& \sqrt{ 25 - \sqrt{432} } \qquad \text{point outside the triangle} \\ x &=& \sqrt{ 4.21539030917} \\ \mathbf{x} &=& \mathbf{2.05314157066} \\ \hline \end{array}\)

\(\mathbf{AB\text{ is } \approx 2.05 }\)

A circle has center \(O\) and radius \(6\).
Points \(A,\ B,\ C,\ \) and \(D\) are chosen on the circle such that \(AD=BC\),
and such that rays \(\overrightarrow{AD}\) and \(\overrightarrow{BC}\) meet at point \(P\) outside the circle.
Suppose that \(AB=10\) and \(\angle DOC=90^\circ\). Compute \(AP\).

\(\begin{array}{|rcll|} \hline \text{In $\triangle AEO$ } \\ \hline \mathbf{\sin(A)} &=& \mathbf{\dfrac{5}{6}} \\ \mathbf{A} &=& \mathbf{56.4426902381^\circ} \\ \hline \end{array} \begin{array}{|rcll|} \hline A+B+45^\circ &=& 180^\circ \\ \mathbf{B} &=& \mathbf{135^\circ-A} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{In $\triangle DFO$ } \\ \hline FD&=& 6\sin(45^\circ) \quad | \quad \sin(45^\circ) = \dfrac{\sqrt{2}}{2} \\ FD&=& \dfrac{6\sqrt{2}}{2} \\ \mathbf{FD} &=& \mathbf{3\sqrt{2}} \\ \hline \end{array} \begin{array}{|rcll|} \hline 2C+2\left(\dfrac{B}{2}+45^\circ \right) &=& 180^\circ \\ 2C+B+90^\circ &=& 180^\circ \\ 2C+B&=& 90^\circ \\ 2C &=& 90^\circ-B \quad | \quad B=135^\circ-A \\ 2C &=& 90^\circ-(135^\circ-A) \\ 2C &=& A-45^\circ \\ \mathbf{C} &=& \mathbf{\dfrac{A-45^\circ}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{In $\triangle DFP$ } \\ \hline \sin(C) &=& \dfrac{FD}{x} \quad | \quad \mathbf{FD=3\sqrt{2}} \\\\ \sin(C) &=& \dfrac{3\sqrt{2}}{x} \\\\ x &=& \dfrac{3\sqrt{2}}{\sin(C)} \quad | \quad \mathbf{C=\dfrac{A-45^\circ}{2}} \\\\ x &=& \dfrac{3\sqrt{2}}{\sin\left(\dfrac{A-45^\circ}{2}\right)} \quad | \quad A=56.4426902381^\circ \\\\ x &=& \dfrac{3\sqrt{2}}{\sin\left(\dfrac{ 56.4426902381^\circ-45^\circ}{2}\right)} \\\\ x &=& \dfrac{3\sqrt{2}}{\sin\left(\dfrac{ 11.4426902381^\circ}{2}\right)} \\\\ x &=& \dfrac{3\sqrt{2}}{\sin(5.72134511904^\circ)} \\\\ x &=& \dfrac{3\sqrt{2}}{0.09969044344} \\\\ x &=& \dfrac{4.24264068712}{0.09969044344} \\\\ \mathbf{x} &=& \mathbf{42.5581484121} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{In $\triangle ADO$ } \\ \hline \mathbf{AD} &=& \mathbf{6^2+6^2-2*6*6*\cos(B)} \\\\ AD &=& 72\Big(1 - \cos(B)\Big) \quad | \quad \mathbf{B=135^\circ-A} \\ AD &=& 72\Big(1 - \cos(135^\circ-A)\Big) \\ AD &=& 72\Big(1 - \cos(180^\circ-45^\circ-A)\Big) \\ AD &=& 72\Big(1 - \cos(180^\circ-(45^\circ+A)\Big) \\ AD &=& 72\Big(1 + \cos( 45^\circ+A)\Big) \\ AD &=& 72\Big(1 + \cos(101.442690238^\circ)\Big) \\ \mathbf{AD} &=& \mathbf{57.7160876877} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{AP} &=& \mathbf{x + AD} \\\\ AP &=& 42.5581484121 + 57.7160876877 \\ \mathbf{AP} &=& \mathbf{100.274236100} \\ \hline \end{array}\)

\(\mathbf{AP\approx 100.3} \)

Solve the equation on the interval \([0,2\pi)\) .
\(6\sin^2(x)+3\sin(x)=0\)

What are the correct solutions?

\(\begin{array}{|lrcll|} \hline &\mathbf{6\sin^2(x)+3\sin(x)} &=& \mathbf{0} \quad &| \quad : 3 \\\\ &2\sin^2(x)+ \sin(x) &=& 0 \\ &\underbrace{\sin(x)}_{=0}\Big( \underbrace{2\sin(x)+ 1}_{=0} \Big) &=& 0 \\ \\ \hline 1) & \sin(x) &=& 0 \\ & x &=& \arcsin(0) + \pi n \qquad n \in \mathbb{Z} \\ & x &=& 0 + \pi n \\ & \mathbf{x} &=& \mathbf{\pi n} \qquad n \in \mathbb{Z} \\ \hline 2) & 2\sin(x)+ 1 &=& 0 \\ & \sin(x) &=& -\dfrac{1}{2} \\ & x &=& \arcsin\left(-\dfrac{1}{2}\right) + 2\pi n \qquad n \in \mathbb{Z} \\ & \mathbf{x} &=& \mathbf{-\dfrac{\pi}{6} + 2\pi n} \qquad n \in \mathbb{Z} \\ \hline 3) & \sin(x) &=& -\dfrac{1}{2} \quad & | \quad \sin(x) = \sin(\pi-x) \\ & \sin(\pi-x) &=& -\dfrac{1}{2} \\ & \pi-x &=& \arcsin\left(-\dfrac{1}{2}\right) + 2\pi n \qquad n \in \mathbb{Z} \\ & \pi-x &=& -\dfrac{\pi}{6} + 2\pi n \qquad n \in \mathbb{Z} \\ & x &=& \pi + \dfrac{\pi}{6} + 2\pi m \qquad m \in \mathbb{Z} \\ & \mathbf{x} &=& \mathbf{\dfrac{7\pi}{6}} + 2\pi m \qquad m \in \mathbb{Z} \\ \hline \end{array}\)

On the interval \([0,2\pi)\)

solutions:
\(\begin{array}{|lll|} \hline x &= 0\\\\ x &= \pi\\\\ x &= \dfrac{7 \pi}{6}\\\\ x &= \dfrac{11 \pi}{6} \\ \hline \end{array}\)