2015/119) Prove that $(sin\theta+i\cos\theta)^8=\cos8\theta-i\sin8\theta$

$(\sin\theta+i\cos\theta)^8 = i^8(\cos\theta - i \sin \theta)^8$
$=(e^{-i\theta})^8= e^{-i8\theta}$
$=\cos 8\theta - i \sin 8\theta$

2015/118) Without using logs and calculator find the number of digits in $2^{100}$

$2^{10} = 1024 > 1000 = 10^3$
so $2^{100} > 10^{30}$
further
$2^{10} < 1025 < 10^3 * \frac{41}{40}=  10^3 *( 1+ \frac{1}{40}) $
so $2^{100} < 10^{30} * ( 1+ \frac{1}{40})^{10} < 10^{30} * ( 1+ \frac{1}{40})^{40}<  10^{30} * e <  10^{30} * 3$
so $10^{30} < 2^{100} < 3 * 10^{30}$
hence 31 digits

2015/117) If $m\tan(a-30^\circ)=n\tan(a+120^\circ)$ show that $\cos2a=\frac{m+n}{2(m-n)}$

We have $\tan (a+ 120^\circ) = - \cot(a + 30^\circ)$ using $\tan (x+90^\circ) = - \cot x$
so $m\tan(a-30^\circ)= -n \cot ( a+ 30^\circ)$
so $\tan (a+30^\circ) \tan (a-30^\circ) = \dfrac{-n}{m}$
or $\dfrac{\tan a + \tan 30^\circ}{1- \tan a \tan 30^\circ} * \dfrac{\tan a -\tan 30^\circ}{1 + \tan a \tan 30^\circ)} = \dfrac{-n}{m}$
or $\dfrac{tan ^2 a - tan ^2 30^\circ}{1- tan ^2 a tan ^2 30^\circ} = \dfrac{n}{m}$
or $\dfrac{tan ^2 a - \frac{1}{3}}{1- tan ^2 a \frac{1}{3}} = \dfrac{n}{m}$
or $\dfrac{3 tan ^2 a -1}{3 - tan ^2 a} = \dfrac{n}{m}$
or $\dfrac{1-3 tan ^2 a}{3- tan ^2 a} = \dfrac{n}{m}$
use componendo dividendo to get
$\dfrac{4 - 4 \tan ^2 a}{- 2 - 2 \tan ^2 a} = \dfrac{n+m}{n-m}$
or $2\dfrac{1 - tan ^2 a}{1 + 2 tan ^2 a} = \dfrac{n+m}{m-n}$
or $\dfrac{1 - tan ^2 a}{1 + 2 tan ^2 a} = \dfrac{n+m}{2(m-n)}$
or $\cos 2a = \dfrac{n+m}{2(m-n)}$

2015/116) Find the limit of the product as k goes to infinite $\prod_{n=1}^{k}\frac{n^2}{n^2-1}$

We propose $\prod_{n=1}^{k}\frac{n^2}{n^2-1}=\frac{2k}{k+1}$

for n =2 we have product = $\frac{4}{3} = \frac{2 * 2}{2+1}$

let it be true of n = k

so we have $\prod_{n=1}^{k}\frac{n^2}{n^2-1}=\frac{2k}{k+1}$

multiply by (k+1)st term to get   $f(k+1) = \prod_{n=1}^{k}\frac{n^2}{n^2-1} * \frac{(k+1)^2}{(k+1)^2 - 1}$
= $\frac{2k}{k+1} * \frac{(k+1)^2}{k(k+2)} = \frac{2 * (k +1)}{(k+2)}$
so if it is true for k it is true for k + 1
Now that we have found the closed form as k goes to infinite above product goes to 2 (converges to 2)

2015/115) If $a+b+c=\pi$

Prove that
$\sin^3 a+\sin^3 b+ \sin^3 c=3\cos(\frac{a}{2})\cos(\frac{b}{2})\cos(\frac{c}{2})$
           $+ \cos(\frac{3a}{2})\cos(\frac{3b}{2})\cos(\frac{3c}{2})$

Solution
we have $\sin(3x) = 3\sin(x) - 4\sin^3(x)$
hence
$\sin ^3 x = (\frac{3}{4}\sin x - \frac{1}{4}\sin (3x))$
so 
$\sin^3 A+\sin^3 B + \sin^3 C = \frac{3}{4}( \sin A + \sin B + \sin C) - \frac{1}{4}( \sin 3A + \sin 3B + \sin 3C)\cdots(1)$
 we have if $A + B+ C = \pi$
$\sin \frac{A+B}{2} = sin (\frac{\pi}{2}- \frac{C}{2}) = \cos \frac{C}{2}$
and $\cos \frac{A+B}{2} = \cos (\frac{\pi}{2} - \frac{C}{2}) = \sin  \frac{C}{2}$

Now
$\sin A + \sin B + \sin C= 2 \sin \frac{A+B}{2} \cos\frac{A-B}{2} + 2 \sin \frac{C}{2} \cos \frac{C}{2}$
= $2 \cos\frac{C}{2} \cos\frac{A-B}{2} + 2 \cos\frac{A+B}{2}  \cos \frac{C}{2}$
= $2 \cos   (\cos \frac{A-B}{2} + \cos \frac{A+B}{2})$
= $4 \cos \frac{A}{2}cos \frac{B}{2} \cos \frac{C}{2}$
hence $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2}cos \frac{B}{2} \cos \frac{C}{2}\cdots(2)$
By similar argument if
$3A+3B+3C = 3\pi$
then
$\sin \frac{3A + 3B}{2} = \sin \frac{3\pi- 3c}{2} = \cos \frac{3C}{2}$
and
$\cos \frac{3A + 3B}{2} = \cos \frac{3\pi- 3c}{2} = -\sin \frac{3C}{2}$
using this we get
$\sin 3A + \sin 3B + \sin 3C = - 4 \cos \frac{3A}{2}cos \frac{3B}{2} \cos \frac{3C}{2}\cdots(3)$
using (1), (2),(3) we get the result

2015/114) Solve in real $a^3 + b^3 = 8 – 6ab$

$a^3 + b^3 - 8 + 6ab = 0$

or

$a^3 + b^3 + (-2)^3 -3(-2)(a)(b) = 0$

=> $a + b – 2 = 0$ or $a=b= - 2$

using the fact

$x^3 + y^3 + z^3 – 3xyz = \dfrac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)$

2015/113) There is a number n between to successive squares.this is k larger than the smaller number and l smaller than the larger number . Prove that n-kl is a perfect square.

Let the number n be between $a^2$ and $(a+1)^2$
As per given condition
$n- a^2 = k\cdots(1)$
$(a+1)^2 –n = l\cdots(2)$
Adding (1) and (2)
$k + l = 2a + 1$
or $l = 2a + 1 - k$
now
$n- kl = (a^2+k) – k(2a+1-k)
= a^2 + k – 2ak –k + k^2 = a^2 – 2ak + k^2 = (a-k)^2$
Hence proved

2015/112) show that $\cos2A + \cos6A + \cos8A = \dfrac{\sqrt{13)}-1}{4}$ where $A = \dfrac{pi}{13}$

Let
$x = \cos2A + \cos6A + \cos8A \cdots(1)$
by seeing that $\sqrt{13}$ on right
square both sides of (1) to get
$x^2 = \cos^2 2A + \cos^2 6A + \cos^2 8A + 2 \cos 2A \cos 6A + 2 \cos 2A \cos 8A + 2 \cos 6A \cos 8A$
multiply by 2 to get
$2 x^2 = 2 \cos^2 2A + 2 \cos^2 6A + 2 \cos^2 8A$
        + $2(2 \cos 2A \cos 6A + 2 \cos 2A \cos 8A + 2 \cos 6A \cos 8A)$
= $\cos 4A + 1 + \cos 12 A + 1 + \cos 16 A + 1$ 
                    + $2( \cos 8A +\cos 4 A + \cos 10 A + \cos 6A + \cos 14 A + \cos 2 A)$
= $3 + \cos 4A + \cos 12 A + \cos 16 A$
                      + 2 $( \cos 8A + \cos 6A + \cos 2A + \cos 4A + \cos 10 A + \cos 14A)$

Now $\cos 16 A = \cos 10 A$ as $26 A = 2\pi$
$\cos 14 A = \cos 12 A$ as $26 A = 2\pi$
So we continue
= $3 + \cos 4A + \cos 12 A + \cos 10 A + 2(x+ \cos 4A + \cos 10 A + \cos 12 A)$
= $3 + 2x + 3 (\cos 4A + \cos 12 A+ \cos 10 A)$
Now $\cos 2A + \cos 4A + \cos 6A + \cos 8A + \cos 10 A + \cos 12A = \dfrac{-1}{2}$

So $\cos 4A + \cos 12 A+ \cos 10 A = (\dfrac{- 1}{2}-x)$

So $2x^2 = 3 + 2x + 3(\dfrac{-1}{2}- x)$
Or $4x^2 = 6 + 4x -3 – 6x$
Or $4x^2 + 2x -3 = 0$

This has one positive solution $\dfrac{\sqrt{13)}-1}{4}$ and one negative solution

As $\cos2A + \cos6A + \cos8A = \cos 2A + \cos 6A – \cos 5A$ and $\cos6 A > 0$ and $\cos 2A > \cos 5A$ so this is > 0
So this is $\dfrac{\sqrt{13)}-1}{4}$


I have solved at http://in.answers.yahoo.com/question/index;_ylt=ApwbfmdwbadqGq5DHY.KZsiRHQx.;_ylv=3?qid=20120607073236AAFCjFo

2105/111) Given 4 positive integers a,b,c and d such that $a^5=b^4, c^3=d^2$ and $c−a=19$ what is $d−b$

as $c^3 = d^2$ so c will be a square let $c = x^2$

as $a^5 = b^4$ so $a = y^4$

now
$c-a = 19$
=> $x^2 - y^4 = 19$
=> $(x-y^2)(x+y^2) = 19$
hence $x - y^2 =1$ and $x+y^2 = 19$ as 19 is a prime
so $x = 10$ and $y = 3$

so $a = y^4$ or $a^5 = y^20 = b^ 4$ or $b= y^5 = 243$

$c= x^2 => c= 100$ and hence $d^2 = 10^6$ and so $d = 1000$
$d-b = 1000 - 243 = 757$

2015/110) If 'a' and 'b' are the roots of $x^2-3x+1=0$ then

find the value of $\frac{a^{2014} + b^{2014} + a^{2016} + b^{2016}}{ a^{2015} + b^{2015}}$

Solution

a is root of $x^2 – 3x + 1=0$

so $a^2 - 3a + 1 = 0$

or

$a^2 + 1 = 3a$

so

$\dfrac{a^{2014} + a^{2016}}{a^{2015}} = \dfrac{1+a^2}{a} = 3 \cdots(1)$

Similarly

$\dfrac{b^{2014} + b^{2016}}{b^{2015}} = \dfrac{1+b^2}{b} = 3 \cdots(2)$

using

if $\dfrac{x}{y} = \dfrac{z}{w}$ then both are $\dfrac{x+z}{y+w}$

we get

$\dfrac{a^{2014} + b^{2014} + a ^{2016} + b^{2016}}{a^{2015} + b^{2015}} =  3$

2015/109) A triangle with sides 10, 24, and 26 is inscribed in a circle. What is the radius of the circle?

This is a right angled triangle because $10^2 + 24^2 = 26^2$ ( as u can check)
Therefore the hypotneuse forms the diameter and therefore the radius is $\frac{26}{2} =13$

2015/108) If $a + b = 1$ and $a^2 + b^2 = 2$ what is the value of $a^3 + b^3$ ?

we have

$a+b =1 \cdots(1)$
$a^2+b^2 = 2 \cdots(2)$
from 1st we get
$(a+b)^2 = a^2 +b^2 + 2ab = 2 + 2ab = 1$
or $ab = \dfrac{- 1}{2}$

hence
$a^3+b^3 = (a+b)^3 - 3ab(a+b) = 1 - 3 * \dfrac{-1}{2} * 1 = \dfrac{5}{2}$

2015/107) find the sum of $\frac{1}{3}+\frac{4}{9}+\frac{7}{27}+\frac{10}{81}+\cdots$.

The nth term = $\dfrac{3n+1}{3^{n+1}}$ n is from 0 to infinite
= $\dfrac{n}{3^n}+ \dfrac{1}{3}\cdot\dfrac{1}{3^n}$
The second term is GP and the sum upto infinite terms is $\dfrac{1}{3}\cdot\dfrac{1}{1-\frac{1}{3}} = \dfrac{1}{2}$
let $f(x) = 1 + x + x^2 + \cdots = \dfrac{1}{1-x}$
then differentiate both sides wrt x to get
$f'(x) = 1 + 2x + 3x^2 + \cdots. = \dfrac{1}{(1-x)^2}$
multiply by x to get
$xf'(x) = x + 2x^2 + 3x^3 \cdots = \dfrac{x}{(1-x)^2}$
$x = \dfrac{1}{3}$ gives the 1st sum as $\dfrac{\frac{1}{3}}{(\frac{2}{3})^2}= \dfrac{9}{4}\cdot\dfrac{1}{3} = \dfrac{3}{4}$
so sum = $\dfrac{3}{4}+ \dfrac{1}{2} = \dfrac{5}{4}$