Answer to Question #170798 in Trigonometry for Deniss

Question

Find sin18°.

Answer 1

Let $x=18\degree$

Then:

$2x=90\degree-3x$

$sin2x=sin(90\degree-3x)=cos3x$

$2sinxcosx=4cos^3x-3cosx$

$4cos^2x-2sinx-3=0$

$4(1-sin^2x)-2sinx-3=0$

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

$sinx=sin18\degree=\frac{-2+\sqrt{20}}{8}=\frac{-1+\sqrt{5}}{4}$

$sin18\degree=0.3090$