Question

FIND THE VALUE OF SIN18 DEGREE, SIN36 DEGREE, SIN72 DEGREE, TAN 22 1/2 (22 AND HALF)

Answer 1

Let x= 18 degrees

5x = 90 degrees

3x = 90 - 2x taking sin both side

sin (3x) = sin {(90 - 2x)}

sin 3x = cos 2x

3 sin x - 4 sin^3(x) = 1- 2 sin^2(x)

put sin x = t

4t^3 - 2t^2 - 3t +1 = 0

4 t^3 - 4 t^2 + 2 t^2 -2t -t+1=0

4 t^2(t-1) +2t (t - 1) -1(t-1) =0

(t-1){4t^2 + 2t -1}= 0

either t = 1.........................(i)

or {4t^2 + 2t -1}= 0

t =[ -2 +/- sqrt{4 + 16}]/8

t = { -2 +/- sqrt {20}]/8

t = {-1 + sqrt 5}/4.............(ii) by taking +ve sign

t = {-1 - sqrt 5}/4.............(iii) by taking - ve sign

from (i),(ii) and (iii)

t = sin 18 = 1, {-1 + sqrt 5}/4 and {-1 - sqrt 5}/4

since sin 18 cannot be equal to 1 and {-1 - sqrt 5}/4 being in first quadrant

Therefore sin 18 = {-1 + sqrt 5}/4
I hope this helps u