2022/032) Show that $\sin ^3 18^\circ + \sin ^2 18^\circ = \frac{1}{8}$

we have $\sin\, 54^\circ = \cos \, 36^\circ$

let $x = 18^\circ$

so we have $\sin 3x = \cos 2x$

Or $3 \sin \, x - 4\sin ^3 x = 1 - 2 \sin ^2 x$

putting $\sin \,x = y$ we get

$3y - 4y^3 = 1 - 2y^2$

or $4y^3 - 2y^2 - 3 y + 1 = 0$

or we see that (y-1) is a factor as putting above as f(y) and find f(1) = 0 

by division we get $(y-1)(4y^2 + 2y -1) = 0$

as $\sin\,18^\circ$ is not zero so $4y^2 + 2y - 1=0\cdots(1)$

we need to show $y^3 + y^2 = \frac{1}{8}$

from (1) $4y^2 = - 2y + 1\cdots(2)$

or $8y^3 = 2y(-2y + 1) = - 4y^2 + 2y  = - -4y^2 + 1 - 4y^2$ (from (2))

or $8y^3 + 8y^2 = 1$

or $y^3 + y^2 = \frac{1}{8}$

   proved 

 

2022/031) The number of real roots of the equation $3x^2 - 4 |x^2-1| + x-1 = 0$ is

 there are 2 cases $x^2 >= 1$

this gives $3x^2 - 4(x^2-1) + x -1 = 0$

or $-x^2 + x + 3 = 0$

or $x^2 -x  - 3 = 0$

this has two slutions one in $(1,\infty$ and another is $(-\infty, -1)$ 

for for the other case we have $3x^2 + 4(x^2-1) + x -1 = 0$ or $7x^2 + x - 5 = 0$ f(0) = -5, f(1) = 3 f(-1) = = 1 so there are 2 roots one between 0 and 1 and another between 0 and -1. 

So there are 4 solutions 

2022/030) Factor $x^4+3 x^2 y^2+2 y^4+4 x^2+5 y^2+3$

we have

 $x^4+3 x^2 y^2+2 y^4+4 x^2+5 y^2+3$

$= x^4+3 x^2 y^2+4 x^2 + 2 y^4 +5 y^2+3$ putting them in descending power of x

$=x^4+ x^2(3 y^2 +4) +  (2y^2 + 1)(y^2 + 3)$ factoring the expression involving y only

$=x^4+ x^2((2y^2 + 1) + (y^2 +3) +  (2y^2 + 1)(y^2 + 3)$ split the middle term as $3y^2 + 4 = 2y^2 + 1 + y^3 + 3$

$= x^2(x^2 + 2y^2 + 1) + (y^2 + 3)(x^2 + 2y^2  +1 ) = (x^2 +2y^2 + 1)(x^2 + y^2 + 3)$

2022/029) Show that $\frac{1}{29} + \frac{1}{31} + ... \frac{1}{55}$ when the sum is put in the form $\frac{p}{q}$ then p is divisible by 83

Using $\frac{1}{n} + \frac{1}{a-n} = \frac{a}{n(a-n)}$
We get $\frac{1}{n} =  \frac{a}{n(a-n)}-\frac{1}{a-n}$
Putting a= 83 we get
$\frac{1}{n} =  \frac{83}{n(83-n)}-\frac{1}{83-n}$

Running n from 29 upto 55 in increment of 2 we have

$\sum_{n=14}^{27} \frac{1}{2n+1}=\sum_{n=14}^{27} (\frac{83}{(2n+1)(82-2n)}-\frac{1}{82-2n})$

Or $\sum_{n=14}^{27} \frac{1}{2n+1}=83\sum_{n=14}^{27}\frac{1}{(2n+1)(82-2n)}- \sum_{n=14}^{27}\frac{1}{82-2n}$

As $\sum_{n=14}^{27}\frac{1}{(2n+1)(82-2n)}$ when we take the sum  we shall get it of the form $\frac{p}{q}$ where q does not have a divisor 83. as 83 does not divide any denominator

So  $\sum_{n=14}^{27} \frac{1}{2n+1}=\frac{83p}{q}- \sum_{n=14}^{27}\frac{1}{82-2n}\cdots(1)$

Further $82-2n = 2k => n = 41-k$

So $\sum_{n=14}^{27}\frac{1}{82-2n} = \sum_{n=14}^{27}\frac{1}{2n}\cdots(2)$

 Now the given expression

= $\sum_{n=0}^{27} \frac{1}{2n+1 } -  \sum_{n=1}^{27} \frac{1}{2n}$

= $\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{27} \frac{1}{2n} + \sum_{n=14}^{27} \frac{1}{2n+1 }  $

= $\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{27} \frac{1}{2n} + \frac{83p}{q}- \sum_{n=14}^{27}\frac{1}{82-2n}$

= $\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{27} \frac{1}{2n} + \frac{83p}{q}- \sum_{n=14}^{27}\frac{1}{2n}$

=  $\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{13} \frac{1}{2n} + \frac{83p}{q}- 2 \sum_{n=14}^{27}\frac{1}{2n}$

=  $\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{13} \frac{1}{2n} + \frac{83p}{q}-  \sum_{n=14}^{27}\frac{1}{n}$

=  $\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{13} \frac{1}{2n} + \frac{83p}{q}-   \sum_{n=7}^{13}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{1}{2n+1}$

=  $(\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=7} ^{13} \frac{1}{2n+1}) + \frac{83p}{q}-   \sum_{n=1}^{6}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{2}{2n}$

=  $\sum_{n=0}^{6} \frac{1}{2n+1 } + \frac{83p}{q}-   \sum_{n=1}^{6}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{1}{n}$

= $\frac{83p}{q} + \sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{13} \frac{1}{2n} -   \sum_{n=7}^{13}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{1}{2n+1}$
= $\frac{83p}{q} + \sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=7}^{13} \frac{1}{2n+1} -   \sum_{n=1}^{13}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{1}{2n}$
= $\frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=7}^{13}\frac{1}{2n}-  \sum_{n=7}^{13}\frac{1}{2n}$
= $\frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=7}^{13}\frac{1}{n}$
= $\frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=7}^{13}\frac{1}{n}$
= $\frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=7}^{13}\frac{1}{n}$
= $\frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - (\sum_{n=3}^{6}\frac{1}{2n+1} + \sum_{n=4}^{6}\frac{1}{2n})$
= $\frac{83p}{q} + \sum_{n=0}^{2} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=4}^{6}\frac{1}{2n}$
= $\frac{83p}{q} + \sum_{n=0}^{2} \frac{1}{2n+1 } - \sum_{n=1}^{3}\frac{1}{2n} - \sum_{n=4}^{6}\frac{1}{2n} -  \sum_{n=4}^{6}\frac{1}{2n}$
= $\frac{83p}{q} + \sum_{n=0}^{2} \frac{1}{2n+1 } - \sum_{n=1}^{3}\frac{1}{2n} - \sum_{n=4}^{6}\frac{1}{n} $
= $\frac{83p}{q} + 1 + \frac{1}{3} +  \frac{1}{5} - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} - \frac{1}{4} - \frac{1}{5} - \frac{1}{6}$
$=\frac{83p}{q}$

Numerator is divisible by 83 and denominator is not and 83 is prime . hence proved 

2022/028) Solve $\sin ^5 x + \cos^5 x = 1$

 We know as $-1 \le \sin\, x \le 1$ so $sin^5 x \ le sin ^2 x$. they are same when sin x = 0 or 1 otherwise  $\sin^5 x \lt sin ^2 x$

similarly $\cos^5 x \le \cos ^2 x$. they are same when cos  x = 0 or 1 otherwise  $cos^5 x \ lt cos  ^2 x$

so $\sin ^5 x + \cos^5 x \le \sin ^2 x + \cos^2 x$ or 1 they are same when $\sin \, x =$ 0 or and $\cos\,x$ = 0 or 1 that is one of them is zero and another 1

this is possible when $\sin\, x = 0$ and $\cos\,x = 1$ that is $x = 2n\pi$

or $\sin\, x = 1$ and $\cos\,x = 0$ that is $x = (2n + \frac{1}{2}) \pi$

hence $x \in \{ 2n\pi, (2n + \frac{1}{2}) \pi \} $

2022/027) If $\sum_{k=1}^n \frac{2k+1}{(k^2+k)^2} = .9999$ find n

 We have

$\frac{2k+1}{(k^2+k)^2} = \frac{(k+1)^2 - k^2}{(k+1)^2 k^2} =  \frac{1}{k^2} - \frac{1}{(k+1)^2}$

The above is a telescopic sum and we have

 $\sum_{k=1}^n \frac{2k+1}{(k^2+k)^2} = \frac{1}{1} - \frac{1}{(n+1)^2} = .9999$

so  $\frac{1}{(n+1)^2} = 1- .9999 = .0001 = \frac{1}{100^2}$

n+ 1 = 100 or n = 99