2016/089) Evaluate $\sin\,18^\circ$ and $\cos\,18^\circ$

Let, $A = 18^\circ$
Then $2A = 90^\circ - 3A$
Taking sine on both sides, we get
$\sin 2A = \sin (90^\circ - 3A) = \cos 3A$
$=> 2 \sin\, A \cos\, A = 4 \cos^3 A - 3 \cos\, A$
or $2 \sin\, A \cos\, A - 4 \cos^3 A + 3 \cos\, A = 0$
or $\cos\, A (2 \sin\, A - 4 \cos^2 A + 3) = 0$
Dividing both sides by $\cos\, A = \cos 18^\circ$ which is not zero we get
$2 \sin\, A - 4 (1 - \sin^2 A) + 3 = 0$
or $4 \sin^2 A + 2 \sin\ A - 1 = 0$ which is a quadratic in $\sin\ A$
hence $\sin\,A = \frac{-1\pm\sqrt{5}}{4}$
but as $\sin\, 18^\circ$ is positive we have $\sin 18^\circ = \frac{-1+\sqrt{5}}{4}$
now $\cos^2 18^\circ= 1- \sin ^2 18^\circ = 1 - (\frac{-1+\sqrt{5}}{4})^2$
$= 1 - \frac{5+1-2\sqrt{5}}{16} = \frac{10+2\sqrt{5}}{16}$
 $\cos\,18^\circ= \frac{\sqrt{10+2\sqrt{5}}}{4}$

2016/088) Which number is smaller $\sqrt{3} + \sqrt{5}$ or $\sqrt{2} + \sqrt{6}$

we have $\sqrt{3} - \sqrt{2}= \frac{1}{\sqrt{3} + \sqrt{2}}$
and  $\sqrt{6} - \sqrt{5}= \frac{1}{\sqrt{6} + \sqrt{5}}$
from the above $\sqrt{3} - \sqrt{2} > \sqrt{6} - \sqrt{5}$
or  $\sqrt{3} + \sqrt{5} > \sqrt{6} + \sqrt{2}$

2016/087) If $a\sin\,x=b\sin(x+\frac{2\pi}{3})=c\sin(x+\frac{4\pi}{3})$ prove that $ab+bc+ca=0$

Let $asin\,x=b\sin(x+\frac{2\pi}{3})=c\sin(x+\frac{4\pi}{3})=k$
hence $\frac{k}{a} = \sin\,x\cdots(1)$
$\frac{k}{b} = \sin(x+\frac{2\pi}{3})\cdots(2)$
$\frac{k}{c} = \sin(x+\frac{4\pi}{3})$
or $\frac{k}{c} = \sin(x-\frac{2\pi}{3})\cdots(3)$

from (1),(2) and (3)
$\frac{k}{a} + \frac{k}{b} + \frac{k}{c} =\sin\,x +  \sin(x+\frac{2\pi}{3}) + \sin(x-\frac{2\pi}{3})$
$= \sin\,x +  \sin\,x\cos \frac{2\pi}{3} + \cos \,x\sin  \frac{2\pi}{3} + \sin\,x\cos \frac{2\pi}{3} - \cos \,x\sin  \frac{2\pi}{3}$
$= \sin\,x +  2\sin\,x\cos \frac{2\pi}{3}$
$= \sin\,x +  2\sin\,x( -\frac{1}{2})$
$= \sin\,x -  \sin\,x$
$= 0$

2016/086) If a,b,c are in AP find the fixed point wthough which line $ax+by+c= 0$ passes

a,b,c are in AP so $a+c = 2b$ or $c = 2b-a$
$ax+by+c=0$
$=>ax + by + (2b-a)=0$ or $(x-1) a + (y+2) b=0$
so the point through which the lines pass is (1,-2) as the above equation should be independent of (a,b)