Saved Bookmarks
| 1. |
Sin20.sin40.sin60.sin80=1/16 |
| Answer» since,sin60=√ 3/2= √ 3/2( sin20sin40sin80)=√ 3/2( sin20sin80sin40)=√ 3/4 [(2sin20sin40)sin80]on applying [cos(A-B)-cos(A+B) = 2sinAsinB]we get,= √ 3/4 (cos20-cos60)sin80 [since,cos(-a)=cosa]= √ 3/4(cos20sin80-cos60sin80)= √ 3/8(2sin80cos20-sin80)= √ 3/8(sin100+sin60-sin80)= √ 3/8( √ 3/2+sin100-sin80 )= √ 3/8( √ 3/2+sin(180-80)-sin80 )= √ 3/8( √ 3/2+sin80-sin80 ) [since,sin(180-a)=sina]= √ 3/8( √ 3/2)= 3/16 | |