Question

How do you use the half-angle identity to find the exact value of sin(-105degrees) ?

Answer 1

`(sqrt3+2)/(2sqrt2)`

Explanation:

`sin(-105^@)=-sin105^@`
`-sin(60^@+45^@)`
`=-[sin60^@cos45^@+cos60^@sin45^@]`
`=-[sin(2xx30^@)cos45^@+cos(2xx30^@)sin45^@]`
`=-[2sin30^@cos30^@cos45^2+{cos^2(30)^@-sin^2(30)^@}sin45^@]`
`=-[(2xx1/2xxsqrt3/2xx1/sqrt2)+(3/2-1/2)*1/sqrt2]`
`=-(sqrt3/(2sqrt2)+1/sqrt2)`