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

Question

12471 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-26T04:00:05

It is $sin(18^o)=(-1+sqrt5)/4$

Let $x=18deg=>5x=90deg$

Hence we have that

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

So we have

$cos3x=sin2x$

$=> 4cos^3(x)-3cosx = 2*sinx*cosx$

$=>4cos^2(x)-3 =2sinx$

$=> -4sin^2(x)-2sinx+1 =0$

$=> 4sin^2(x)+2sinx-1 =0$

$=> sinx= (-2+- sqrt(4+16))/8$

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

Hence $sin18^o= (-1+sqrt5)/4$