How do you find the exact value of Sin (18 Degrees)?

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Answer 1 · 2015-09-26T04:00:05

It is $sin (18^{o}) = \frac{- 1 + \sqrt{5}}{4}$ $sin(18^o)=(-1+sqrt5)/4$

Let $x = 18 d e g \Rightarrow 5 x = 90 d e g$ $x=18deg=>5x=90deg$

Hence we have that

$cos 3 x = sin (90 - 3 x) = sin (5 x - 3 x)$ $cos3x=sin(90-3x)=sin(5x-3x)$

So we have

$cos 3 x = sin 2 x$ $cos3x=sin2x$

$\Rightarrow 4 {cos}^{3} (x) - 3 cos x = 2 \cdot sin x \cdot cos x$ $=> 4cos^3(x)-3cosx = 2*sinx*cosx$

$\Rightarrow 4 {cos}^{2} (x) - 3 = 2 sin x$ $=>4cos^2(x)-3 =2sinx$

$\Rightarrow - 4 {sin}^{2} (x) - 2 sin x + 1 = 0$ $=> -4sin^2(x)-2sinx+1 =0$

$\Rightarrow 4 {sin}^{2} (x) + 2 sin x - 1 = 0$ $=> 4sin^2(x)+2sinx-1 =0$

$\Rightarrow sin x = \frac{- 2 \pm \sqrt{4 + 16}}{8}$ $=> sinx= (-2+- sqrt(4+16))/8$

Since 18deg is in the 1st quadrant, we take only the positive value.

Hence ${sin 18}^{o} = \frac{- 1 + \sqrt{5}}{4}$ $sin18^o= (-1+sqrt5)/4$