Question

`Sin(40°)+cos(50°)+sin(80°)-cos(190°) =` ?

Answer 1

`Sin(40°)+cos(50°)+sin(80°)-cos(190°)`

`=Sin(40°)+cos(90^@-40°)+sin(80°)-cos(270^@-80^@)`

`=Sin(40°)+sin(40°)+sin(80°)+sin(80^@)`

`=2(Sin(40°)+sin(80°))`

`=4sin60^@cos20^@=sqrt3cos20^@`