by solving the equation sin2A=cos3A. find the value of sin18 .show your ans. in the form (sq root a)+b/c where a,c,b are natural nos.

Dear yashika

given

sin2A=cos3A

2 sinA cosA=4cos³A-3cosA

2 sin A=4 cos²A-3

2 sin A =4(1-sin²A)-3

now solve for sinA

sin A=(-1-√5)/4  or (-1+√5)/4

Put A=18

and for 0<A<180 so value of sin will be positive

sin18=(-1+√5)/4