Question

The value of sin ${18}^0$

• A. $\frac{\sqrt{5}-1}{4}$
• B. $\frac{\sqrt{5}+1}{4}$
• C. $-\frac{\sqrt{5}+1}{4}$
• D. $\frac{1-\sqrt{5}}{4}$

Answer 1

The correct option is A $\frac{\sqrt{5}-1}{4}$

$cos{18}^0=sin({90}^0-{18}^0)=sin{72}^0$
$=2sin{36}^{0}cos{36}^0$
$=2\left(2sin{18}^{0}cos{18}^0\right)(co{s}^2{18}^0-si{n}^2{18}^0)$
$=4\left(sin{18}^{0}cos{18}^0\right)(1-2si{n}^2{18}^0)$
$cos{18}^0=4\left(sin{18}^{0}cos{18}^0\right)(1-2si{n}^2{18}^0)$
$1=4sin{18}^0(1-2si{n}^2{18}^0)$
$8si{n}^3{18}^0-4sin{18}^0+1=0$
The following cubic equation is divisible by $sin{18}^0=\frac{1}{2}$
Therefore
$(2sin{18}^0-1)(8sin{18}^0{}^2+4sin{18}^0-2)=0$
$2(2sin{18}^0-1)(4sin{18}^0{}^2+2sin{18}^0-1)=0$
Using quadratic formula we get
$2(2sin{18}^0-1)(4sin{18}^0+(\sqrt{5}+1\left)\right)(4sin{18}^0-(\sqrt{5}-1\left)\right)=0$
$sin{18}^0=\frac{1}{2},sin{18}^0=-\frac{\sqrt{5}+1}{4},sin{18}^0=\frac{\sqrt{5}-1}{4}$
$sin\theta$ is positive in the first quadrant and $sin{30}^0=\frac{1}{2}$
Therefore $sin{18}^0=\frac{\sqrt{5}-1}{4}$
Hence answer is A