Grade 9 Extra exercise (Algebraic expression )

 

Exercise:

Given that,

$A\left( x \right)={{\left( 5x-1 \right)}^{2}}-{{\left( 2x+3 \right)}^{2}}$ and $B\left( x \right)=9{{x}^{2}}-24x+16-\left( 4-3x \right)\left( 3x+5 \right)$

1)    Expand and reduce $A\left( x \right)$

2)    Solve $A\left( x \right)=-8$

3)    Show that, $A\left( x \right)=\left( 7x+2 \right)\left( 3x-4 \right)$

4)    Factorize $B\left( x \right)$

5)    Given that,

$H\left( x \right)=\frac{A\left( x \right)}{B\left( x \right)}$

a)    For what values of $x$ is $H$ defined

b)    Simplify $H\left( x \right)$

c)    Solve $H\left( x \right)=2$

Solution:

1)    ${{\left( 5x-1 \right)}^{2}}={{\left( 5x \right)}^{2}}-2\left( 5x \right)\left( 1 \right)+1=25{{x}^{2}}-10x+1$

${{\left( 2x+3 \right)}^{2}}={{\left( 2x \right)}^{2}}+2\left( 2x \right)\left( 3 \right)+{{3}^{2}}=4{{x}^{2}}+12x+9$

So that, $A\left( x \right)=25{{x}^{2}}-10x+1-\left( 4{{x}^{2}}+12x+9 \right)=25{{x}^{2}}-10x+1-4{{x}^{2}}-12x-9$

                    $=\left( 25-4 \right){{x}^{2}}+\left( -10-12 \right)x+1-9=21{{x}^{2}}-22x-8$

2)    $A\left( x \right)=-8\Leftrightarrow 21{{x}^{2}}-22x=0\Leftrightarrow x\left( 21x-22 \right)=0\Leftrightarrow x=0\,\,or\,\,x=\frac{22}{21}$

3)    $A\left( x \right)=\left[ \left( 5x-1 \right)-\left( 2x+3 \right) \right]\left[ \left( 5x-1 \right)+\left( 2x+3 \right) \right]$

        $=\left( 5x-1-2x-3 \right)\left( 5x-1+2x+3 \right)=\left( 3x-4 \right)\left( 7x+2 \right)$

Or we can expand it and compare

4)    $B\left( x \right)=9{{x}^{2}}-24x+16-\left( 4-3x \right)\left( 3x+5 \right)$

Notice that, $9{{x}^{2}}-24x+16={{\left( 3x \right)}^{2}}-2\left( 3x \right)\left( 4 \right)+{{4}^{2}}={{\left( 3x-4 \right)}^{2}}$

So that, $B\left( x \right)={{\left( 3x-4 \right)}^{2}}-\left( 4-3x \right)\left( 3x+5 \right)={{\left( 3x-4 \right)}^{2}}+\left( 3x-4 \right)\left( 3x+5 \right)$

                     $=\left( 3x-4 \right)\left( 3x-4+3x+5 \right)=\left( 3x-4 \right)\left( 6x+1 \right)$

5)     a) $H$ is defined when $B\left( x \right)\ne 0\Leftrightarrow \left( 3x-4 \right)\left( 6x+1 \right)\ne 0$ i.e $x\ne \frac{4}{3}$ nor $x\ne \frac{-1}{6}$

b)$H\left( x \right)=\frac{A\left( x \right)}{B\left( x \right)}=\frac{\left( 3x-4 \right)\left( 7x+2 \right)}{\left( 3x-4 \right)\left( 6x+1 \right)}=\frac{7x+2}{6x+1}$

c) $H\left( x \right)=2\Leftrightarrow \frac{7x+2}{6x+1}=2\Leftrightarrow 7x+2=12x+2\Leftrightarrow x=0$

Grade 9 Extra Exercise (Radicals)

Exercise : Compute the following,

1)    $\sqrt{{{3}^{2}}+{{4}^{2}}+{{\left( 2\sqrt{6} \right)}^{2}}}$

2)    $\sqrt{\frac{{{8}^{10}}+{{4}^{10}}}{{{8}^{4}}+{{4}^{11}}}}$

3)    $\frac{5\sqrt{12}+\sqrt{48}}{2\sqrt{3}-\sqrt{192}-\sqrt{27}}$

4)    $\frac{{{\left( 10\sqrt{3} \right)}^{4}}}{25\times {{10}^{3}}\times 3+6\times 37.5\times {{10}^{3}}}$

5)    $\frac{0.48\times {{\left( {{10}^{3}} \right)}^{4}}\times 0.001}{0.3\times {{10}^{-4}}\times 100}$

6)    $\sqrt{{{\left( 2\sqrt{5} \right)}^{2}}-\sqrt{{{2}^{3}}+\sqrt{{{10}^{2}}-{{6}^{2}}}}}$

7)    $\sqrt{6-\sqrt{\frac{7}{2}+\frac{\sqrt{12}+\sqrt{27}}{\sqrt{300}}}}$

Solution:

1)    $\sqrt{{{3}^{2}}+{{4}^{2}}+{{\left( 2\sqrt{6} \right)}^{2}}}=\sqrt{9+16+{{2}^{2}}\times \sqrt{{{6}^{2}}}}=\sqrt{25+4\times 6}=\sqrt{25+24}=\sqrt{49}=\sqrt{{{7}^{2}}}=7$

2)    $\sqrt{\frac{{{8}^{10}}+{{4}^{10}}}{{{8}^{4}}+{{4}^{11}}}}=\sqrt{\frac{{{\left( 2\times 4 \right)}^{10}}+{{4}^{10}}}{{{\left( 2\times 4 \right)}^{4}}+{{4}^{11}}}}=\sqrt{\frac{{{2}^{10}}\times {{4}^{10}}+{{4}^{10}}}{{{2}^{4}}\times {{4}^{4}}+{{4}^{4+7}}}}=\sqrt{\frac{{{4}^{10}}\left( {{2}^{10}}+1 \right)}{{{4}^{4}}\left( {{2}^{4}}+{{4}^{7}} \right)}}$

                 $=\sqrt{\frac{{{4}^{10}}}{{{4}^{4}}}\times \frac{{{2}^{10}}+1}{{{2}^{4}}+{{\left( {{2}^{2}} \right)}^{7}}}}=\sqrt{{{4}^{10-4}}\times \frac{{{2}^{10}}+1}{{{2}^{4}}+{{2}^{14}}}}=\sqrt{{{4}^{6}}\times \frac{{{2}^{10}}+1}{{{2}^{4}}\left( 1+{{2}^{10}} \right)}}$

                 $=\sqrt{\frac{{{4}^{6}}}{{{2}^{4}}}}=\sqrt{\frac{{{\left( {{2}^{2}} \right)}^{6}}}{{{2}^{4}}}}=\sqrt{\frac{{{2}^{12}}}{{{2}^{4}}}}=\sqrt{{{2}^{12-4}}}=\sqrt{{{2}^{8}}}=\sqrt{{{\left( {{2}^{4}} \right)}^{2}}}={{2}^{4}}=16$

3)    Prime decomposition for $12={{2}^{2}}\times 3$, $48={{2}^{4}}\times 3$, $192={{2}^{6}}\times 3$ and $27=3\times {{3}^{2}}={{3}^{3}}$

$\frac{5\sqrt{12}+\sqrt{48}}{2\sqrt{3}-\sqrt{192}-\sqrt{27}}=\frac{5\sqrt{{{2}^{2}}\times 3}+\sqrt{{{2}^{4}}\times 3}}{2\sqrt{3}-\sqrt{{{2}^{6}}\times 3}-\sqrt{{{3}^{3}}}}=\frac{5\left( \sqrt{{{2}^{2}}}\times \sqrt{3} \right)+\sqrt{{{2}^{4}}}\times \sqrt{3}}{2\sqrt{3}-\sqrt{{{2}^{6}}}\times \sqrt{3}-\sqrt{{{3}^{2}}}\times \sqrt{3}}$

                           \[=\frac{\left( 5\times 2 \right)\sqrt{3}+\sqrt{{{2}^{2}}\times {{2}^{2}}}\times \sqrt{3}}{2\sqrt{3}-\sqrt{{{\left( {{2}^{3}} \right)}^{2}}}\times \sqrt{3}-3\sqrt{3}}=\frac{10\sqrt{3}+\sqrt{{{2}^{2}}}\times \sqrt{{{2}^{2}}}\times \sqrt{3}}{2\sqrt{3}-{{2}^{3}}\sqrt{3}-3\sqrt{3}}=\frac{10\sqrt{3}+4\sqrt{3}}{2\sqrt{3}-8\sqrt{3}-3\sqrt{3}}\]

\[=\frac{14\sqrt{3}}{\left( 2-8-3 \right)\sqrt{3}}=\frac{14\sqrt{3}}{-9\sqrt{3}}=-\frac{14}{9}\]

 

4)    We have, \[{{\left( 10\sqrt{3} \right)}^{4}}={{10}^{4}}\sqrt{{{3}^{4}}}={{10}^{4}}\sqrt{{{\left( {{3}^{2}} \right)}^{2}}}={{3}^{2}}\times {{10}^{4}}=9\times {{10}^{4}}\]

So that, \[\frac{{{\left( 10\sqrt{3} \right)}^{4}}}{25\times {{10}^{3}}\times 3+6\times 37.5\times {{10}^{3}}}=\frac{9\times {{10}^{4}}}{25\times {{10}^{3}}\times 3+225\times {{10}^{3}}}\]

$=\frac{9\times {{10}^{4}}}{{{10}^{3}}\left( 25\times 3+9\times 25 \right)}=\frac{9\times 10}{3\left( 25+3\times 25 \right)}=\frac{3\times 10}{25\left( 1+3 \right)}=\frac{3\times 2}{5\times 4}=\frac{3}{5\times 2}=\frac{3}{10}$

5)    $\frac{0.48\times {{\left( {{10}^{3}} \right)}^{4}}\times 0.001}{0.3\times {{10}^{-4}}\times 100}=\frac{0.48\times {{10}^{12}}\times {{10}^{-3}}}{3\times {{10}^{-1}}\times {{10}^{-4}}\times {{10}^{2}}}=\frac{48\times {{10}^{-2}}\times {{10}^{9}}}{3\times {{10}^{-5}}\times {{10}^{2}}}=\frac{48\times {{10}^{7}}}{3\times {{10}^{-3}}}=1.6\times {{10}^{11}}$

6)    $\sqrt{{{\left( 2\sqrt{5} \right)}^{2}}-\sqrt{{{2}^{3}}+\sqrt{{{10}^{2}}-{{6}^{2}}}}}=\sqrt{{{\left( 2\sqrt{5} \right)}^{2}}-\sqrt{{{2}^{3}}+\sqrt{100-36}}}=\sqrt{{{\left( 2\sqrt{5} \right)}^{2}}-\sqrt{{{2}^{3}}+\sqrt{64}}}$

$=\sqrt{{{\left( 2\sqrt{5} \right)}^{2}}-\sqrt{{{2}^{3}}+8}}=\sqrt{{{\left( 2\sqrt{5} \right)}^{2}}-\sqrt{16}}=\sqrt{{{\left( 2\sqrt{5} \right)}^{2}}-4}=\sqrt{4\times 5-4}=\sqrt{16}=4$

7)     We have $\sqrt{12}=\sqrt{{{2}^{2}}\times 3}=\sqrt{{{2}^{2}}}\times \sqrt{3}=2\sqrt{3}$,    

      $\sqrt{27}=\sqrt{{{3}^{2}}\times 3}=\sqrt{{{3}^{2}}}\times \sqrt{3}=3\sqrt{3}$ and $\sqrt{300}=\sqrt{3\times 100}=\sqrt{3}\times \sqrt{100}=10\sqrt{3}$

So that, \[\sqrt{6-\sqrt{\frac{7}{2}+\frac{\sqrt{12}+\sqrt{27}}{\sqrt{300}}}}=\sqrt{6-\sqrt{\frac{7}{2}+\frac{5\sqrt{3}}{10\sqrt{3}}}}=\sqrt{6-\sqrt{\frac{7}{2}+\frac{1}{2}}}=\sqrt{6-\sqrt{4}}\]

                                              \[=\sqrt{6-2}=\sqrt{4}=2\]

Nice exercise with nice tech

 Exercise:

Determine the values of $a\And b$ such that ${{\left( \frac{1+\sqrt{5}}{2} \right)}^{12}}=a+b\sqrt{5}$

Solution:

Let $u=\frac{1+\sqrt{5}}{2}\Leftrightarrow 2u=1+\sqrt{5}\Leftrightarrow 2u-1=\sqrt{5}$

Squaring both sides to get $4{{u}^{2}}-4u+1=5\Leftrightarrow 4\left( {{u}^{2}}-u-1 \right)=0\Leftrightarrow $

Notice that, ${{u}^{3}}={{u}^{2}}+u=2u+1\And {{u}^{6}}={{\left( {{u}^{3}} \right)}^{2}}={{\left( 2u+1 \right)}^{2}}=4{{u}^{2}}+4u+1$

                                                                $=4\left( u+1 \right)+4u+1=8u+5$

It follows that , ${{u}^{12}}={{\left( {{u}^{6}} \right)}^{2}}={{\left( 8u+5 \right)}^{2}}=64{{u}^{2}}+25+80u=64\left( u+1 \right)+25+80u$

                          $=64u+80u+64+25=144u+89$

So that, ${{\left( \frac{1+\sqrt{5}}{2} \right)}^{12}}=\frac{144\left( 1+\sqrt{5} \right)}{2}+89=72\left( 1+\sqrt{5} \right)+89=161+72\sqrt{5}=a+b\sqrt{5}$

Therefore, after comparing we obtain $a=161\And b=72$

Nice exercise -Secondary Classes -Exams

Exercise:

Solve in $\mathbb{R}$,

1)      $\sqrt{x}-\left| x-2 \right|<1$

2)      ${{\sin }^{4}}x+{{\cos }^{4}}x=1$

3)      $\sum\nolimits_{i=0}^{5}{{{\left( -1 \right)}^{i+1}}{{x}^{i}}=0}$

Solution:

1)      We know that, if $\sqrt{f\left( x \right)}<g\left( x \right)$ then $g\left( x \right)>0$

So that, $1+\left| x-2 \right|>0\Leftrightarrow \left| x-2 \right|>-1$  true for all $x$

Notice that, the domain of this equation $x\ge 0$ 

So that there are 2 cases  considered in this study

Case 1: $x>2$

$\sqrt{x}-\left| x-2 \right|<1\Leftrightarrow \sqrt{x}-x+2<1\Leftrightarrow \sqrt{x}-x+1<0\Leftrightarrow {{x}^{2}}-3x+1>0$

So that, $x>\frac{3+\sqrt{5}}{2}$

Case 2: $x<2$

$\sqrt{x}-\left| x-2 \right|<1\Leftrightarrow \sqrt{x}+x-2<1\Leftrightarrow \sqrt{x}+x-3<0\Leftrightarrow {{x}^{2}}-7x+9>0$

So that, $0\le x<\frac{7-\sqrt{13}}{2}$

2)      We have ${{\sin }^{4}}x+{{\cos }^{4}}x=1\Leftrightarrow {{\left( {{\sin }^{2}}x \right)}^{2}}+{{\left( {{\cos }^{2}}x \right)}^{2}}=1$

$\Leftrightarrow {{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}=1+2{{\sin }^{2}}x{{\cos }^{2}}x\Leftrightarrow 2{{\sin }^{2}}x{{\cos }^{2}}x=0$  

$\Leftrightarrow \sqrt{2}\sin x\cos x=0\Leftrightarrow \sin x=0\,\,or\,\,\,\cos x=0\Leftrightarrow x=0\,\,or\,\pi +2k\pi \,\,or\,\,x=\pm \frac{\pi }{2}+2k\pi \,\,,\,\,k\in \mathbb{Z}$

Or we can use factorization without going complete squaring as follows

${{\sin }^{4}}x+{{\cos }^{4}}x=1\Leftrightarrow {{\cos }^{4}}x=1-{{\sin }^{4}}x=\left( 1-{{\sin }^{2}}x \right)\left( 1+{{\sin }^{2}}x \right)={{\cos }^{2}}x\left( 1+{{\sin }^{2}}x \right)$

so that, ${{\cos }^{2}}x\left[ {{\cos }^{2}}x-\left( 1+{{\sin }^{2}}x \right) \right]=0\Leftrightarrow {{\cos }^{2}}x\left( \cos 2x-1 \right)=0$

$\Leftrightarrow x=\pm \frac{\pi }{2}+2k\pi \,\,or\,\,x=0+k\pi $ with $k\in \mathbb{Z}$

3)      $S=\sum\nolimits_{i=0}^{5}{{{\left( -1 \right)}^{i+1}}{{x}^{i}}=-1+x-{{x}^{2}}+{{x}^{3}}-{{x}^{4}}+{{x}^{5}}}=0$

Note that, the above series is geometric with common ratio is $-x$ and initial value $-1$

So that, $S=\sum\nolimits_{i=0}^{n-1}{a{{r}^{i}}}=-\left( \frac{1-{{\left( -x \right)}^{6}}}{1+x} \right)=0\Leftrightarrow 1-{{x}^{6}}=0\Leftrightarrow {{x}^{6}}=1\Leftrightarrow x=1$ 

Nice exercise to review Riemann sum

Exercise:

Evaluate the following $\underset{n\to \infty }{\mathop{\lim }}\,\frac{\sqrt[{{n}^{2}}]{\prod\nolimits_{k=1,2,...,n}{k!}}}{\sqrt{n}}$ 

Solution: Let $f\left( n \right)=\frac{\sqrt[{{n}^{2}}]{\prod\nolimits_{k=1,2,..,n}{k!}}}{\sqrt{n}}=\frac{{{\left( \prod\nolimits_{k=1,2,..,n}{k!} \right)}^{1/{{n}^{2}}}}}{{{n}^{1/2}}}$ 

So that, $\ln f\left( n \right)=\frac{1}{{{n}^{2}}}\ln \left( \prod\nolimits_{k=1,2,..,n}{k!} \right)-\frac{1}{2}\ln n$
Notice that, $g\left( n \right)=\ln \left( \prod\nolimits_{k=1,2,..,n}{k!} \right)=\sum\nolimits_{k=1,2,..,n}{\ln k!}=\ln 1!+\ln 2!+...+\ln n!$
Moreover, we know that, $\ln n!=\ln \left( n\left( n-1 \right)\times ....2\times 1 \right)=\ln n+\ln \left( n-1 \right)+...+\ln 2+\ln 1=\sum\nolimits_{k=1}^{n}{\ln k}$
So that, $g\left( n \right)=\underbrace{\ln 1}_{\ln 1!}+\underbrace{\ln 1+\ln 2}_{\ln 2!}+\underbrace{\ln 3+\ln 2+\ln 1}_{\ln 3!}+....+\underbrace{\sum\nolimits_{k=1}^{n}{\ln k}}_{\ln n!}$ 
                  $=n\ln 1+\left( n-1 \right)\ln 2+\left( n-2 \right)\ln 3+....+\ln n$
Hence $g\left( n \right)=n\ln 1-n\ln n+n\ln n+\left( n-1 \right)\ln 2-\left( n-1 \right)\ln n+\left( n-1 \right)\ln n+...+\ln n-\ln n+\ln n$       
   $=n\left( \ln 1-\ln n \right)+\left( n-1 \right)\left( \ln 2-\ln n \right)+....+\ln \left( n/n \right)+\frac{n\left( n+1 \right)}{2}\ln n$
Thus, $\frac{1}{{{n}^{2}}}g\left( n \right)=\frac{1}{n}\ln \left( 1/n \right)+\frac{n-1}{{{n}^{2}}}\ln \left( 2/n \right)+....+\frac{1}{{{n}^{2}}}\ln \left( n/n \right)+\frac{n+1}{2n}\ln n$
                    $=\frac{1}{n}\left( \ln \frac{1}{n}+\frac{n-1}{n}\ln \frac{2}{n}+.....+\frac{1}{n}\ln \frac{n}{n} \right)+\frac{n+1}{2n}\ln n$
So that, $\ln f\left( n \right)=\frac{1}{n}\sum\nolimits_{k=1,..,n}{\frac{n-\left( k-1 \right)}{n}\ln \frac{k}{n}}+\frac{n+1}{2n}\ln n-\frac{1}{2}\ln n$
                       $=\frac{1}{n}\sum\nolimits_{k=1,..,n}{\frac{n-\left( k-1 \right)}{n}\ln \frac{k}{n}+\frac{1}{2n}\ln n}$ 
Hence $\underset{n\to \infty }{\mathop{\lim }}\,\ln f\left( n \right)={{I}_{n}}+\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{2n}\ln n$ where ${{I}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\nolimits_{k=1}^{n}{\frac{n-\left( k-1 \right)}{n}\ln \frac{k}{n}}$ 
Notice that, this sum is Riemann sum with $\Delta x=\frac{b-a}{n}=\frac{1-0}{n}$ and ${{x}_{i}}=a+i\Delta x$
Hence our partition will be like $0={{x}_{0}}<{{x}_{1}}<{{x}_{2}}<....<{{x}_{n}}=1$
Moreover, $\frac{n-k+1}{n}\ln \left( k/n \right)=\left( 1-\frac{k}{n}+\frac{1}{n} \right)\ln \left( k/n \right)=\left( 1-{{x}_{i}} \right)f\left( {{x}_{i}} \right)\Delta {{x}_{i}}$
Therefore, ${{I}_{n}}=\int_{0}^{1}{\left( 1-x \right)\ln x\,dx}=-\frac{3}{4}$ and $\underset{n\to \infty }{\mathop{\lim }}\,\frac{\ln n}{2n}=0$ 
Therefore, $\underset{n\to \infty }{\mathop{\lim }}\,f\left( n \right)={{e}^{-3/4}}$  

Equation with radicals asked by Husni Elayyan Madarseh Madarseh in math group معلمو الرياضيات محافظة قلقيلية

Exercise:
Solve in $\mathbb{R}$, $x=\sqrt{21-\sqrt{21+x}}$
Solution: In order to crack this equation we will use zero trick
So that, $x=\sqrt{21+x-x-\sqrt{21+x}}=\sqrt{{{\left( \sqrt{21+x} \right)}^{2}}-\left( x+\sqrt{21+x} \right)}$
Put $w=\sqrt{21+x}$ then $x=\sqrt{{{w}^{2}}-\left( x+w \right)}\Leftrightarrow {{x}^{2}}={{w}^{2}}-\left( x+w \right)$
i.e $\left( x-w \right)\left( x+w \right)+\left( x+w \right)=0\Leftrightarrow \left( x+w \right)\left( x-w+1 \right)=0$
it follows that, $w=-x\,\,\,or\,\,\,x=w-1$
so that, $\sqrt{21+x}-1=x\Leftrightarrow {{\left( x+1 \right)}^{2}}=21+x\Leftrightarrow {{x}^{2}}+2x+1=21+x$
$\Rightarrow {{x}^{2}}+x-20=0\Leftrightarrow \left( x+5 \right)\left( x-4 \right)=0$
Thus the accepted roots for this equation is $x=4$

Nice exercise include all levels for high school

Exercise:

Show that, the ratio of AB over BC is golden ratio if an only

if the quadrilateral of side $AB$ is square in the right figure.


Solution: Suppose that, the circle is unit circle then $r=1$

On the other hand we have the two triangles $NCM$ and $MBC$

are similar by A.A due to the following reasons

We have $\angle M=90$ (inscribed angle property), then $\angle M=\angle B$ and $\angle C$ is common angle

thus, the similar ratio with $BC=x\And BM=y$

 $\frac{NCM}{MCB}=\frac{NC}{MC}=\frac{CM}{CB}=\frac{NM}{MB}=\frac{2r}{MC}=\frac{MC}{y}=\frac{NM}{x}\Leftrightarrow M{{C}^{2}}=2ry=2y$

From Pythagoras theorem in triangle$MBC$,$M{{C}^{2}}={{x}^{2}}+{{y}^{2}}=2y$    (*)

Notice that, $\angle MKC=\angle MNC=\frac{\overset\frown{MC}}{2}$ and x$\angle KMN=\angle KCN=\frac{\overset\frown{NK}}{2}$

But $\left( KM \right)\parallel \left( AB \right)$ as $KMBA$ is rectangle, then by alternate interior

Angles property we have $\angle MKC=\angle KCN$ and $\angle KMN=\angle MNC$

Thus $\angle MNC=\angle KCN$ i.e  $\vartriangle QNC$ is isosceles triangle of vertex $Q$      (**)

Also we have $\angle NKC=\angle NMC=90$ (inscribed angle property) and $NC$ is common side

Thus $\vartriangle NKC$ and $\vartriangle NMC$ are equal by $H.R.A$ therefore, $NK=MC\And NM=KC$

In addition to that, $\angle KNA=\frac{KM+MC}{2}\And \angle MCB=\frac{KM+NK}{2}\Leftrightarrow \angle MCB=\angle KNA$
Also $AK=MB=x$ hence $\vartriangle NKA$ and $\vartriangle MBC$ are equal by $H.R.A$

therefore, $NA=BC$ and $NC=2y+AB\Leftrightarrow AB=2\left( 1-y \right)$             (***)

Now if $KMBA$ is square, then $AB=BM\Leftrightarrow x=2\left( 1-y \right)\Leftrightarrow M{{C}^{2}}={{\left( 2\left( 1-y \right) \right)}^{2}}+{{y}^{2}}=2y$

$\Rightarrow 4\left( 1-2y+{{y}^{2}} \right)+{{y}^{2}}-2y=0\Leftrightarrow 5{{y}^{2}}-10y+4=0\Leftrightarrow y=\frac{5\pm \sqrt{5}}{5}$

$\Leftrightarrow \frac{AB}{BC}=\frac{2\left( 1-y \right)}{y}=\frac{\mp \frac{2}{\sqrt{5}}}{\frac{5\pm \sqrt{5}}{5}}=\frac{\mp 10}{5\sqrt{5}\pm 5}=\frac{\mp 2}{\sqrt{5}\pm 1}\times \frac{\sqrt{5}\mp 1}{\sqrt{5}\mp 1}=\frac{\mp 2\left( \sqrt{5}\mp 1 \right)}{4}=\frac{1\mp \sqrt{5}}{2}$

But $AB>0\And BC>0$ thus $\frac{AB}{BC}=\frac{1+\sqrt{5}}{2}>0$ is only the acceptable ratio

However, if $KMBA$ is rectangle then $AB\ne BM$ as result we have on the other hand

$\vartriangle NCW$ is isosceles triangle of vertex $W$ hence, $\left( OW \right)$ is the perpendicular

bisector of $\left[ NC \right]$ this means that, $Q$ is the orthocenter of this triangle $NCW$ and

$QA=QB$ hence $O$ is the midpoint of $\left[ AB \right]$.from orthocenter property we have

$QM=\frac{NM}{3}\And NQ=\frac{2NM}{3}$

Hence, From Thales’s we have $\frac{MQ}{QN}=\frac{KQ}{QC}=\frac{KM}{NC}$ with $QN=QC\And QK=QM$ then

$\Leftrightarrow \frac{QM}{QN}=\frac{KM}{NC}\Leftrightarrow \frac{1}{2}=\frac{KM}{2}\Leftrightarrow $ hence, $AB=KM=1$ and $2y+1=2\Leftrightarrow y=0.5$

Therefore, $\frac{AB}{BC}=\frac{1}{y}=2$ impossible as $AB<NC$ therefore the accepted case for $ABKM$

must be square inorder to obtain golden ratio.                          Q.E.D
.

Math joke for i love math

$Exercise\,\, 1:\,\,Solve\,\, the\,\, following\,\, inequality,\\

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\log\sqrt[t]{\frac{i}{3m}}<\frac{h}{\log_{a}{10}}\\

Solution:\,\, we\,\, know\,\, that,\,\,\sqrt[t]{w}=\left ( w \right )^{\frac{1}{t}},then\\


\log\sqrt[t]{\frac{i}{3m}}=\log\left ( \frac{i}{3m} \right )^{\frac{1}{t}}
=\frac{1}{t}\log\left ( \frac{i}{3m} \right )<\frac{h}{\log_{a}{10}}

\\

but \log_{a}{10}=\frac{\log 10}{\log a}\,\, then \frac{h}{\log_{a}10} =\frac{h}{\frac{\log10}{\log a}}=h\log a \\

hence,\,\,\log\left ( \frac{i}{3m} \right )<\log a^{th}\\


Thus,\,\,i<3\,\,ma^{th}$

Nice and wonderful exercise shared by Dan Sitrau in RMM facebook group

Exercise:

Show that, $\int_{0}^{\gamma }{\ln \left( 1+\tan \gamma \tan x \right)dx}=\gamma \ln \sec \gamma $ with $-\frac{\pi }{2}\le \gamma \le \frac{\pi }{2}$

Solution: Let $I=\int_{0}^{\gamma }{\ln \left( 1+\tan \gamma \tan x \right)dx}=\int_{0}^{\gamma }{\ln \left( 1+\tan \gamma \tan \left( \gamma -x \right) \right)dx}$

(          using $\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx}$ ……….(*)           )

But $\tan \left( \gamma -x \right)=\frac{\tan \gamma -\tan x}{1+\tan x\tan \gamma }\Leftrightarrow \tan \gamma \tan \left( \gamma -x \right)=\frac{-\tan x\tan \gamma +{{\tan }^{2}}\gamma }{1+\tan x\tan \gamma }$

Hence, $I=\int_{0}^{\gamma }{\ln \left( 1+\frac{-\tan x\tan \gamma +{{\tan }^{2}}\gamma }{1+\tan x\tan \gamma } \right)dx}=\int_{0}^{\gamma }{\ln \left( \frac{1+\tan x\tan \gamma -\tan x\tan \gamma +{{\tan }^{2}}\gamma }{1+\tan x\tan \gamma } \right)dx}$

$\Rightarrow I=\int_{0}^{\gamma }{\ln \left( 1+{{\tan }^{2}}\gamma  \right)dx-}\int_{0}^{\gamma }{\ln \left( 1+\tan x\tan \gamma  \right)dx}$

$\Rightarrow 2I=\int_{0}^{\gamma }{\ln \left( 1+{{\tan }^{2}}\gamma  \right)dx}=\int_{0}^{\gamma }{\ln \left( {{\sec }^{2}}\gamma  \right)dx}=2\gamma \ln \sec \gamma $

Therefore, $I=\int_{0}^{\gamma }{\ln \left( 1+\tan \gamma \tan x \right)dx}=\gamma \ln \sec \gamma $                               Q.E.D



(*) The idea of solution credit to the professor Ravi Prakash

Nice puzzle posted by art of mathematics facebook page

Exercise:

Compute, $\int_{0}^{1}{{{e}^{{{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{{{e}^{x}}}}}}}}}}}}{{e}^{{{e}^{{{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{x}}}}}}}}}}}}\cdots {{e}^{{{e}^{{{e}^{x}}}}}}{{e}^{{{e}^{x}}}}{{e}^{x}}}dx$

Solution: Let’s find the general formula using induction procedure as follows

Clearly, $\int{{{e}^{{{e}^{x}}}}{{e}^{x}}}dx=\int{{{e}^{{{e}^{x}}}}d\left( {{e}^{x}} \right)}={{e}^{{{e}^{x}}}}+c$

Let $u={{e}^{x}}\Leftrightarrow du={{e}^{x}}dx$ , then

$\int{{{e}^{{{e}^{{{e}^{x}}}}}}{{e}^{{{e}^{x}}}}{{e}^{x}}}dx=\int{{{e}^{{{e}^{u}}}}{{e}^{u}}}du=\int{{{e}^{{{e}^{u}}}}d\left( {{e}^{u}} \right)}={{e}^{{{e}^{u}}}}+c={{e}^{{{e}^{{{e}^{x}}}}}}+c$

Thus,$\int{{{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{{{e}^{x}}}}}}}}}}\cdots {{e}^{{{e}^{x}}}}{{e}^{x}}}dx={{e}^{{{e}^{{{}^{{{e}^{{{e}^{x}}}}}}}}}}+c$

Therefore,$\int_{0}^{1}{{{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{{{e}^{x}}}}}}}}}}\cdots {{e}^{{{e}^{x}}}}{{e}^{x}}}dx=\left( {{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{{{e}^{x}}}}}}}}}} \right)_{0}^{1}={{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{{{e}^{1}}}}}}}}}}-{{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{{{e}^{0}}}}}}}}}}={{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{{{e}^{e}}}}}}}}-{{e}^{{{e}^{{{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}^{e}}}}}}<\infty $

Nice exercise posted by Karam Makhlof Knama in their facebook wall

Exercise:

Solve in $\mathbb{R}$, $\sqrt{x+4+2\sqrt{x+3}}+3\sqrt{x+4-2\sqrt{x+3}}=\left( x+2 \right)\sqrt{x+3}$

Solution: we have \[x+4+2\sqrt{x+3}=x+3+1+2\left( 1 \right)\sqrt{x+3}={{\left( \sqrt{x+3}+1 \right)}^{2}}\] and

$x+4-2\sqrt{x+3}=x+3+1-2\left( 1 \right)\sqrt{x+3}={{\left( \sqrt{x+3}-1 \right)}^{2}}$ ,then

$\sqrt{{{\left( \sqrt{x+3}+1 \right)}^{2}}}+3\sqrt{{{\left( \sqrt{x+3}-1 \right)}^{2}}}=\left( x+2 \right)\sqrt{x+3}$

However, $\frac{1}{\sqrt{x+3}+1}\times \frac{\sqrt{x+3}-1}{\sqrt{x+3}-1}=\frac{\sqrt{x+3}-1}{x+3-1}=\frac{\sqrt{x+3}-1}{x+2}$

$\Rightarrow \left( \sqrt{x+3}+1 \right)\left( \sqrt{x+3}-1 \right)=x+2$

$\sqrt{{{\left( \sqrt{x+3}+1 \right)}^{2}}}+3\sqrt{{{\left( \frac{x+2}{\sqrt{x+3}+1} \right)}^{2}}}=\left( x+2 \right)\sqrt{x+3}$

$\Rightarrow \sqrt{{{\left( \sqrt{x+3}+1 \right)}^{2}}}+3\left( x+2 \right)\sqrt{\frac{1}{{{\left( \sqrt{x+3}+1 \right)}^{2}}}}-\left( x+2 \right)\sqrt{x+3}=0$

Put $y=\sqrt{x+3}>0$ then ${{y}^{2}}-3=x$

$\Rightarrow \sqrt{{{\left( y+1 \right)}^{2}}}+3\left( {{y}^{2}}-3+2 \right)\frac{1}{\sqrt{{{\left( y+1 \right)}^{2}}}}-\left( {{y}^{2}}-3+2 \right)y=0$

$\Rightarrow y+1+\frac{3{{y}^{2}}-3}{y+1}-{{y}^{3}}+y=0$ multiply both sides by $y+1$ to get

$\Rightarrow {{\left( y+1 \right)}^{2}}+3\left( {{y}^{2}}-1 \right)-y\left( {{y}^{2}}-1 \right)\left( y+1 \right)=0\Leftrightarrow \left( y+1 \right)\left[ y+1+3\left( y-1 \right)-y\left( {{y}^{2}}-1 \right) \right]=0$

$\Rightarrow y=-1$ rejected or $y+1+3y-3-{{y}^{3}}+y=0\Leftrightarrow -{{y}^{3}}+5y-2=0$

\[\Rightarrow {{y}^{3}}-y-4y+2{{y}^{2}}-2{{y}^{2}}+2=0\]

$\Rightarrow {{y}^{3}}-y+2{{y}^{2}}-4y-2{{y}^{2}}+2=y\left( {{y}^{2}}+2y-1 \right)-2\left( {{y}^{2}}+2y-1 \right)=0$

$\Rightarrow \left( {{y}^{2}}+2y-1 \right)\left( y-2 \right)=0\Leftrightarrow {{y}^{2}}+2y-1=0\,\,\,or\,\,\,y=2$

But ${{y}^{2}}+2y+1-1-1={{\left( y+1 \right)}^{2}}=2\Leftrightarrow y=-1\pm \sqrt{2}$ but $y=-1-\sqrt{2}<0$ rejected

If $y=2\Leftrightarrow x+3=4\Leftrightarrow$ $x=1$. However, $y=-1\pm \sqrt{2}$ with

$x=-2\sqrt{2}$ is rejected after substation in the original equation

By parts integrations applications

Exercise:

Consider the sequence ${{\left( {{I}_{n}} \right)}_{n>0}}$ is defined to be ${{I}_{n}}=\int_{1}^{e}{x{{\left( \ln x \right)}^{n}}dx}$

Show that, ${{e}^{2}}=2{{I}_{n}}+n{{I}_{n-1}}$ for every $n\in \mathbb{N}$

Solution: We have${{I}_{n}}=\int_{1}^{e}{x{{\left( \ln x \right)}^{n}}dx}$

Let $u={{\left( \ln x \right)}^{n}}\And dv=x\Leftrightarrow du=\frac{n}{x}{{\left( \ln x \right)}^{n-1}}\And v=\frac{{{x}^{2}}}{2}$ , then

${{I}_{n}}=\int_{1}^{e}{x{{\left( \ln x \right)}^{n}}dx}=\left[ \frac{{{x}^{2}}}{2}{{\left( \ln x \right)}^{n}} \right]_{1}^{e}-\int_{1}^{e}{\frac{{{x}^{2}}}{2}\frac{n}{x}{{\left( \ln x \right)}^{n-1}}dx}=\frac{{{e}^{2}}}{2}-\frac{n}{2}\int_{1}^{e}{x{{\left( \ln x \right)}^{n-1}}dx}$

Hence, $2{{I}_{n}}={{e}^{2}}-n{{I}_{n-1}}\Leftrightarrow {{e}^{2}}=2{{I}_{n}}-n{{I}_{n-1}}$                                  Q.E.D



--------------------------------------------------------------------------------------------------------------------------


Exercise:

Consider the sequence ${{\left( {{J}_{n}} \right)}_{n}}$ defined to be ${{J}_{n}}=\int_{0}^{\pi /2}{{{\sin }^{n}}x}\,dx$

Show that, for all natural numbers $n$ we have $\left( n+2 \right){{J}_{n+2}}=\left( n+1 \right){{J}_{n}}$

Solution: we have ${{J}_{n+2}}=\int_{0}^{\pi /2}{{{\sin }^{n+2}}x\,dx}=\int_{0}^{\pi /2}{{{\sin }^{n+1+1}}xdx}=\int_{0}^{\pi /2}{{{\sin }^{n+1}}x\sin xdx}$

Let $u={{\sin }^{n+1}}x\And dv=\sin x\Leftrightarrow du=\left( n+1 \right){{\sin }^{n}}x\cos xdx\And v=-\cos x$

${{J}_{n+2}}=\int_{0}^{\pi /2}{{{\sin }^{n+2}}dx}=-\left( {{\sin }^{n+1}}x\cos x \right)_{0}^{\pi /2}+\int_{0}^{\pi /2}{\cos x\left( n+1 \right){{\sin }^{n}}x\cos xdx}$

       $=\left( n+1 \right)\int_{0}^{\pi /2}{{{\sin }^{n}}x{{\cos }^{2}}xdx}=\left( n+1 \right)\int_{0}^{\pi /2}{{{\sin }^{n}}x\left( 1-{{\sin }^{2}}x \right)dx}$

       $=\left( n+1 \right)\int_{0}^{\pi /2}{{{\sin }^{n}}xdx-\left( n+1 \right)\int_{0}^{\pi /2}{{{\sin }^{n+2}}x\,dx}}$

       $=\left( n+1 \right){{J}_{n}}-\left( n+1 \right){{J}_{n+2}}$

Thus, $\left( n+1+1 \right){{J}_{n+2}}=\left( n+1 \right){{J}_{n}}\Leftrightarrow \left( n+2 \right){{J}_{n+2}}=\left( n+1 \right){{J}_{n}}$

Nice and easy exercise asked in كوكب الرياضيات facebook group

Exercise:

Solve in $\mathbb{R}$, $\sqrt[3]{x+10}-\sqrt[3]{x}=1$

Solution: Let $a=\sqrt[3]{x+10}$ and $b=\sqrt[3]{x}$ then $a-b=1$

But ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$

\(\Rightarrow 1=x+10-x-3ab\left( a-b \right)\Leftrightarrow 9-3ab\left( a-b \right)=0\)

So that, $ab\left( a-b \right)=3\Leftrightarrow ab=3\Leftrightarrow \sqrt[3]{x\left( x+10 \right)}=3\Leftrightarrow x\left( x+10 \right)=27$

${{x}^{2}}+10x-27=0\Leftrightarrow {{\left( x+5 \right)}^{2}}=52$ hence, $x=-5\pm 2\sqrt{13}$

If $x<0$ then $\sqrt[3]{x+10}<\sqrt[3]{10}\approx 2.15$ i.e $3=\sqrt[3]{x\left( x+10 \right)}<0$ which is impossible

Thus, $x=-5+2\sqrt{13}$ is the only acceptable real root for this equation.               Q.E.D

Nice exercise posted by me collected from different math bloggers

Exercise:

Evaluate the following,

 $\int_{0}^{1}{\left( \sqrt{x\sqrt[3]{x\sqrt[4]{x......}}}+\sqrt{x+\sqrt{x+\sqrt{x+.....}}} \right)dx}$

Solution:

We have, $\int_{0}^{1}{\left( \sqrt{x\sqrt[3]{x\sqrt[4]{x......}}}+\sqrt{x+\sqrt{x+\sqrt{x+.....}}} \right)dx}$

           $=\int_{0}^{1}{\sqrt{x\sqrt[3]{x\sqrt[4]{x.....}}}dx}+\int_{0}^{1}{\sqrt{x+\sqrt{x+\sqrt{x+.....}}}dx}$

           $={{A}_{1}}+{{A}_{2}}$

Notice that, $\sqrt{x\sqrt[3]{x}}={{\left( x{{\left( x \right)}^{1/3}} \right)}^{1/2}}={{x}^{1/2}}.{{x}^{1/6}}={{x}^{\frac{1}{2!}+\frac{1}{3!}}}$

Thus, we may generate this result to be

                 $\sqrt{x\sqrt[3]{x\sqrt[4]{x.....}}}={{\left( x{{\left( x{{\left( x \right)}^{1/4}}.... \right)}^{1/3}} \right)}^{1/2}}={{x}^{\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+....}}$

Remember that, ${{e}^{x}}=\sum\nolimits_{n=0}^{\infty }{\frac{{{x}^{n}}}{n!}}=1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+....$

Hence $e=2+\frac{1}{2!}+\frac{1}{3!}+....\Leftrightarrow \sum\nolimits_{n=2}^{\infty }{\frac{1}{n!}=e-2}$

Thus, ${{A}_{1}}=\int_{0}^{1}{\sqrt{x\sqrt[3]{x\sqrt[4]{x......}}}dx}=\int_{0}^{1}{{{x}^{e-2}}dx}=\left[ \frac{{{x}^{e-1}}}{e-1} \right]_{0}^{1}=\frac{1}{e-1}$

To compute ${{A}_{2}}$ we need to start by defining a recursive sequence ${{\left( {{f}_{n}}\left( x \right) \right)}_{n\ge 1}}$

with ${{f}_{1}}\left( x \right)=\sqrt{x}$ and ${{f}_{n+1}}\left( x \right)=\sqrt{x+{{f}_{n}}\left( x \right)}$ for all $0\le x\le 1$

We have ${{f}_{n+1}}\left( x \right)=\sqrt{x+{{f}_{n}}\left( x \right)}<{{f}_{n}}\left( x \right)$ hence ${{f}_{n}}\left( x \right)>{{f}_{n+1}}\left( x \right)$.

So that, ${{\left( {{f}_{n}} \right)}_{n}}$ is a monotone increasing sequence bounded above by $x>0$.

By the monotone convergent theorem${{f}_{n}}\to f$ as $n\to \infty $ hence

It follows that, $f\left( x \right)=\sqrt{x+f\left( x \right)}$ then ${{f}^{2}}\left( x \right)-f\left( x \right)+x=0$

$\Rightarrow {{f}^{2}}\left( x \right)-2\frac{1}{2}f\left( x \right)+\frac{1}{4}-\frac{1}{4}-x=0\Leftrightarrow {{\left( f\left( x \right)-\frac{1}{2} \right)}^{2}}=\frac{1+4x}{4}\Leftrightarrow f\left( x \right)=\frac{1}{2}\pm \sqrt{x+\frac{1}{4}}$

Now we are not sure which sign is corrected one

For any $0<x\le 1$ ,
we have $\frac{1}{4}<x+\frac{1}{4}\le 1+\frac{1}{4}\Leftrightarrow \frac{1}{2}<\sqrt{x+\frac{1}{4}}\le \frac{\sqrt{5}}{2}\Leftrightarrow 0<-\frac{1}{2}+\sqrt{x+\frac{1}{4}}\le \frac{\sqrt{5}}{2}\Rightarrow 0<-{{f}_{1}}\left( x \right)$

but $f\left( x \right)>0$ hence the accepted root is $f\left( x \right)=\frac{1}{2}+\sqrt{x+\frac{1}{4}}$

it follows that, ${{A}_{2}}=\int_{0}^{1}{\underset{n\to \infty }{\mathop{\lim }}\,{{f}_{n}}\left( x \right)dx}=\int_{0}^{1}{f\left( x \right)dx}=\int_{0}^{1}{\frac{1}{2}+\sqrt{x+\frac{1}{4}}\,dx}$

Let $u=x+\frac{1}{4}\Leftrightarrow du=dx$ and $u\left( 0 \right)=\frac{1}{4}\And u\left( 1 \right)=\frac{5}{4}$

Hence, \[{{A}_{2}}=\int_{0}^{1}{f\left( x \right)dx}=\left[ \frac{1}{2}x \right]_{0}^{1}+\int_{0}^{1}{\sqrt{x+\frac{1}{4}}\,dx}=\frac{1}{2}+\int_{1/4}^{5/4}{\sqrt{u}\,du}=\frac{1}{2}+\frac{2}{3}\left( {{u}^{3/2}} \right)_{1/4}^{5/4}=\frac{5}{12}\left( 1+\sqrt{5} \right)\]

Nice Exercise asked in many facebook math groups

Exercise

Show that, $\sin 6\sin 54\sin 66=\frac{\sqrt{5}-1}{16}$


Solution:

We know that, $\sin a\sin b=\frac{1}{2}\left( \cos (a-b)-\cos (a+b\right) )$ ,then

$\sin 6\sin 66=\frac{1}{2}\left( \cos (6-66\right) -\cos (6+66))$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\left( \cos \left( -60\right)
-\cos \left( 72\right) \right) $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\left( \cos 60-\cos 72\right) $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\left( \frac{1}{2}-\cos 72\right)
=\frac{1}{4}-\frac{1}{2}\cos 72$

hence, $\sin 6\sin 54\sin 66=\frac{1}{4}\sin 54-\frac{1}{2}\sin 54\cos 72$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =%
\frac{1}{4}\sin \left( 90-36\right) -\frac{1}{2}\sin \left( 90-36\right)
\cos \left( 90-18\right) $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1%
}{4}\cos 36-\frac{1}{2}\cos 36\sin 18$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1%
}{2}\cos 36\left( \frac{1}{2}-\sin 18\right) $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1%
}{2}\cos 36\left( \frac{1}{2}-\frac{\sqrt{5}-1}{4}\right) $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1%
}{2}\cos 36\left( \frac{2-\sqrt{5}+1}{4}\right) $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{%
2}\cos 36\left( \frac{3-\sqrt{5}}{4}\right) $

but $\sin 54=\sin \left( 3\times 18\right) ~$\ and $\sin \left( 5\times
18\right) =\sin 90=1$

however, $\sin 36=\sin \left( 90-54\right) =\cos 54$

Let $\theta =18$ then $\ \sin 2\theta =\cos 3\theta $

but $\cos 3\theta =4\cos ^{3}\theta -3\cos \theta $

hence, $\sin 2\theta =4\cos ^{3}\theta -3\cos \theta \Leftrightarrow 2\cos
\theta \sin \theta =4\cos ^{3}\theta -3\cos \theta $

$\Rightarrow \cos \theta \left( 2\sin \theta -4\cos ^{2}\theta +3\right) =0$

$\Rightarrow \cos \theta =0$ \ or \ $2\sin \theta -4\left( 1-\sin ^{2}\theta
\right) +3=0$

but $\theta \neq 0,2\pi $ hence, $2\sin \theta -4+4\sin ^{2}\theta +3=0$

$\Leftrightarrow \left( 2\sin \theta \right) ^{2}+2\sin \theta
-1=0\Leftrightarrow \left( 2\sin \theta \right) ^{2}+\frac{1}{2}\left(
2\right) \left( 1\right) \left( 2\sin \theta \right) +\frac{1}{4}-\frac{1}{4}%
-1=0$

$\Leftrightarrow \left( 2\sin \theta +\frac{1}{2}\right) ^{2}=\frac{1}{4}+1=%
\frac{5}{4}\Leftrightarrow 2\sin \theta =\frac{-1\pm \sqrt{5}}{2}%
\Leftrightarrow \sin \theta =\frac{-1\pm \sqrt{5}}{4}....(1)$

but $\sin 18\in \left[ 0,90\right] ~~$hence $\sin 18=\frac{\sqrt{5}-1}{4}>0$

However, $\cos 36=\cos \left( 2\theta \right) =1-2\sin ^{2}\theta =1-2\left(
\frac{\sqrt{5}-1}{2}\right) ^{2}=\frac{\sqrt{5}+1}{4}$

Therefore, $A=\sin 6\sin 54\sin 66=\frac{1}{2}\cos 36\left( \frac{3-\sqrt{5}%
}{4}\right) $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\sqrt{5}+1}{8}\times \frac{%
3-\sqrt{5}}{4}=\frac{2\sqrt{5}-2}{32}=\frac{\sqrt{5}-1}{16}$

Trigonometric exercise asked by Dr. Yousef abbas in theمنبر الرياضيات facebook group

Exercise:

Evaluate the following, $\cos \frac{\pi }{7}\cos \frac{2\pi }{7}%
\cos \frac{4\pi }{7}$

Solution:

we have A=$\cos \theta \cos 2\theta \cos 4\theta $ then

$\sin \theta A=\sin \theta \cos \theta \cos 2\theta \cos 4\theta $


$2\sin \theta A=\sin 2\theta \cos 2\theta \cos 4\theta $

$4\sin \theta A=\sin 4\theta \cos 4\theta $

$8\sin \theta A=\sin 8\theta $

thus, $A=\frac{\sin 8\theta }{8\sin \theta }$

but $\sin 8\theta =\sin \left( \frac{8\pi }{7}\right) =\sin \left( \pi +%
\frac{\pi }{7}\right) =-\sin \left( \frac{\pi }{7}\right) =-\sin \theta $

Therefore, $A=\frac{-\sin \theta }{8\sin \theta }=-\frac{1}{8}$

Nice Exercise shared in Art of mathematics page

Exercise:

Evaluate, $\int_{0}^{1}{\frac{12dx\left( {{\left( dx \right)}^{2}}-1 \right)}{\left( 1+\frac{1}{dx} \right)\left( dx-1 \right)\left( d\left( {{x}^{3}}+3x \right) \right)}=??}$

Solution: we have \[d\left( {{x}^{3}}+3x \right)=\left( 3{{x}^{2}}+3 \right)dx\] and\[{{\left( dx \right)}^{2}}-1=\left( dx-1 \right)\left( dx+1 \right)\] then

$\int_{0}^{1}{\frac{12dx\left( {{\left( dx \right)}^{2}}-1 \right)}{\left( 1+\frac{1}{dx} \right)\left( dx-1 \right)\left( d\left( {{x}^{3}}+3x \right) \right)}=\int_{0}^{1}{\frac{12dx\left( dx+1 \right)}{\left( 1+\frac{1}{dx} \right)\left( 3{{x}^{2}}+3 \right)dx}}}$

\[=\int_{0}^{1}{\frac{12dx}{\frac{\left( 3{{x}^{2}}+3 \right)dx}{dx}}=\int_{0}^{1}{\frac{12}{3{{x}^{2}}+3}dx}=4\int_{0}^{1}{\frac{dx}{{{x}^{2}}+1}}=4\left( \arctan x \right)_{0}^{1}=\pi }\]

Nice exericse asked by Dan Sitaru in many math groups

Exercise:

Let ${{\left( {{a}_{n}} \right)}_{n\ge 1}}$ be recursive sequence defined in such way

${{a}_{1}}=a\in \mathbb{R}\backslash \mathbb{Q}$ and $2{{a}_{n}}{{a}_{n+1}}=4{{a}_{n}}+3{{a}_{n+1}}$ for all $n\ge 1$

1) Find the recursive closed formula for ${{a}_{n}}$

2) find$\underset{n\to \infty }{\mathop{\lim }}\,{{a}_{n}}$

Solution:

1) we have $2{{a}_{n}}{{a}_{n+1}}=4{{a}_{n}}+3{{a}_{n+1}}\Leftrightarrow \frac{4{{a}_{n}}+3{{a}_{n+1}}}{{{a}_{n}}{{a}_{n+1}}}=2\Leftrightarrow \frac{4}{{{a}_{n+1}}}+\frac{3}{{{a}_{n}}}=2$

notice that, $2=\frac{8+6}{7}=\frac{4\times 2}{7}+\frac{2\times 3}{7}$  , then

$4\left[ \frac{1}{{{a}_{n+1}}}-\frac{2}{7} \right]+3\left[ \frac{1}{{{a}_{n}}}-\frac{2}{7} \right]=0\Leftrightarrow \left( \frac{1}{{{a}_{n+1}}}-\frac{2}{7} \right)+\frac{3}{4}\left( \frac{1}{{{a}_{n}}}-\frac{2}{7} \right)=0$

Put ${{w}_{n}}=\frac{1}{{{a}_{n}}}-\frac{2}{7}\Leftrightarrow {{w}_{n+1}}=\frac{1}{{{a}_{n+1}}}-\frac{2}{7}\Leftrightarrow {{w}_{n+1}}=-\frac{3}{4}{{w}_{n}}$

Note that, for $n=1$ we have

                       $\begin{align}
  & {{w}_{2}}=\left( -\frac{3}{4} \right){{w}_{1}} \\
 & {{w}_{3}}=\left( \frac{-3}{4} \right){{w}_{2}} \\
 & \,\,\,\,\,\,\,\vdots \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
 & {{w}_{n+1}}=\left( \frac{-3}{4} \right){{w}_{n}} \\
\end{align}$

Hence ${{w}_{n+1}}=\underbrace{\left( \frac{-3}{4} \right)\left( \frac{-3}{4} \right)....\left( \frac{-3}{4} \right)}_{n-time}{{w}_{1}}\Leftrightarrow {{w}_{n+1}}={{\left( \frac{-3}{4} \right)}^{n}}{{w}_{1}}\,\,\forall n\ge 1$

Therefore, $\left( \frac{1}{{{a}_{n+1}}}-\frac{2}{7} \right)={{\left( \frac{3}{4} \right)}^{n}}\left( \frac{1}{{{a}_{1}}}-\frac{2}{7} \right)\Leftrightarrow \frac{1}{{{a}_{k}}}={{\left( \frac{3}{4} \right)}^{k-1}}\left( \frac{1}{a}-\frac{2}{7} \right)+\frac{2}{7}\,\,\,\,\,\forall k\ge 2$

2) as $n\to \infty $ then ${{\left( \frac{-3}{4} \right)}^{n-1}}\to 0$  hence $\frac{1}{{{a}_{n}}}\to \frac{2}{7}$ thus $\underset{n\to \infty }{\mathop{\lim }}\,{{a}_{n}}=\frac{7}{2}=3.5$

*___________________

idea of solution credit to professor Ravi Prakash

Nice exercise asked by Kunihiko Chikaya in Enjoy math group

Exercise: <Japan – Level>

Let $n\in \mathbb{Z}_{+}^{*}$ and let ${{\left( {{f}_{n}} \right)}_{n\ge 0}}$ be a sequence of functions defined over $\left( 0,1 \right)$ in such way ${{f}_{n}}\left( t \right)=\frac{1}{2t}\int_{0}^{1}{\left| {{x}^{n}}-t \right|dx}+\frac{1}{2}$ with ${{a}_{n}}$ is the minimum of ${{f}_{n}}$

Show that, $\underset{n\to \infty }{\mathop{\lim }}\,\prod\limits_{n\le i\le 2n-1}{{{a}_{i}}}=\frac{1}{{{2}^{\ln 2}}}$

Solution: from the definition of absolute value we have

$\left| {{x}^{n}}-t \right|=\left\{ \begin{align}
  & {{x}^{n}}-t\,\,,\,\,\,{{t}^{1/n}}\le x<1 \\
 & 0\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,\,\,x={{t}^{1/n}} \\
 & t-{{x}^{n}}\,\,\,,\,\,\,\,0<x<{{t}^{1/n}} \\
\end{align} \right.$

Hence,${{f}_{n}}\left( t \right)=\frac{1}{2t}\int_{0}^{{{t}^{1/n}}}{\left( t-{{x}^{n}} \right)dx}+\frac{1}{2t}\int_{{{t}^{1/n}}}^{1}{\left( {{x}^{n}}-t \right)dx}+\frac{1}{2}$

               $=\frac{1}{2t}\left[ \left( tx \right)_{0}^{{{t}^{1/n}}}-\frac{1}{n+1}\left( {{x}^{n+1}} \right)_{0}^{{{t}^{1/n}}} \right]+\frac{1}{2t}\left[ \frac{1}{n+1}\left( {{x}^{n+1}} \right)_{{{t}^{1/n}}}^{1}-\left( tx \right)_{{{t}^{1/n}}}^{1} \right]+\frac{1}{2}$

               $=\frac{1}{2t}\left[ {{t}^{1/n+1}}-\frac{1}{n+1}{{t}^{\frac{n+1}{n}}} \right]+\frac{1}{2t}\left[ \frac{1}{n+1}\left( 1-{{t}^{\frac{n+1}{n}}} \right)-\left( t-{{t}^{1/n+1}} \right) \right]+\frac{1}{2}$

              $=\frac{1}{2t}\left[ {{t}^{1/n+1}}-\frac{{{t}^{\frac{n+1}{n}}}}{n+1}+\frac{1}{n+1}-\frac{{{t}^{\frac{n+1}{n}}}}{n+1}-t+{{t}^{1/n+1}} \right]+\frac{1}{2}$

    $=\frac{1}{2t}\left[ 2t.{{t}^{1/n}}-\frac{2}{n+1}{{\left( {{t}^{n+1}} \right)}^{1/n}}+\frac{1}{n+1}-t \right]+\frac{1}{2}={{t}^{1/n}}-\frac{1}{\left( n+1 \right)}\left( {{t}^{1/n}} \right)+\frac{1}{2t\left( n+1 \right)}-\frac{1}{2}+\frac{1}{2}$

    $={{t}^{1/n}}\left( 1-\frac{1}{n+1} \right)+\frac{1}{2t\left( n+1 \right)}=\frac{n}{n+1}{{t}^{1/n}}+\frac{1}{2t\left( n+1 \right)}=\frac{1}{n+1}\left[ n{{t}^{1/n}}+\frac{1}{2t} \right]$

Hence ${{f}_{n}}\left( t \right)=\frac{1}{n+1}\left( n{{t}^{1/n}}+1/2t \right)$ but ${{a}_{n}}$ is the minimum of ${{f}_{n}}$

Hence ${{f}_{n}}'\left( t \right)=\frac{1}{n+1}\left( {{t}^{1/n-1}}-\frac{1}{2{{t}^{2}}} \right)\Leftrightarrow {{f}_{n}}'\left( t \right)=0\Leftrightarrow {{t}^{1/n-1}}=\frac{1}{2{{t}^{2}}}\Leftrightarrow t={{\left( \frac{1}{2} \right)}^{1/n+1}}$

So that ${{f}_{n}}$ attend a local max or min at $t={{\left( 1/2 \right)}^{1/n+1}}$

But ${{f}_{n}}''\left( t \right)=\frac{1}{n+1}\left( \left( 1/n-1 \right){{t}^{1/n-2}}+\frac{1}{{{t}^{3}}} \right)$

We have $0<t<1\Leftrightarrow 0<{{t}^{1/n-2}}<1\And 0<{{t}^{3}}<1\Leftrightarrow {{t}^{-3}}>1$ hence ${{f}_{n}}''\left( t \right)>0$

This proves that ${{f}_{n}}$ has local minimum \[{{a}_{n}}=\underset{0<t<1}{\mathop{\min }}\,{{f}_{n}}\left( t \right)={{f}_{n}}\left( {{\left( 1/2 \right)}^{1/n+1}} \right)={{\left( 1/2 \right)}^{\frac{1}{n+1}}}\]

Let $L=\underset{n\to \infty }{\mathop{\lim }}\,\prod\limits_{n\le i\le 2n-1}{{{a}_{i}}\Leftrightarrow \ln L=\ln }\left( \underset{n\to \infty }{\mathop{\lim }}\,\prod\limits_{n\le i\le 2n-1}{{{a}_{i}}} \right)=\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{i=n}^{2n-1}{\ln {{a}_{i}}}$

Notice that, $\ln {{a}_{n}}=\frac{1}{n+1}\ln \left( 1/2 \right)=-\frac{\ln 2}{n+1}$ hence

$\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{i=n}^{2n-1}{\ln {{a}_{i}}}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{i=n}^{2n-1}{\frac{\ln {{2}^{-1}}}{i}}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{i=1}^{n}{\frac{\ln {{2}^{-1}}}{n+i}}=-\ln 2\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{k=1}^{n}{\frac{1}{n+k}}$

Notice that, $A=\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{k=1}^{n}{\frac{1}{n+k}}$ is Riemann sum with $\Delta {{x}_{i}}=\frac{k}{n}$ and $h=\frac{1-0}{n}$

In partition $0={{x}_{0}}<{{x}_{1}}<{{x}_{2}}<....<{{x}_{n}}=1$ and ${{x}_{i}}=a+ih$

Hence $\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{k=1}^{n}{\frac{1}{n+k}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\nolimits_{k=1}^{n}{\frac{1}{1+k/n}=\int_{0}^{1}{\frac{1}{x+1}dx}=\ln \left| x+1 \right|_{0}^{1}=\ln 2}}$

Thus $\ln L=-\ln 2\ln 2=\ln {{2}^{-\ln 2}}\Leftrightarrow L=\frac{1}{{{2}^{\ln 2}}}$

Idea of solution Credit to Dr. B. Barakat