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)=(-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 #