Question

Given cos(t)=-7/9 and sin(t)= -sqrt32/9, both in the 3rd Quadrant how do you find cos(t/2) and sin(t/2)?

Answer 1

`cos(t/2)=-1/3`
`sin(t/2)=(2sqrt2)/3`

Explanation:

A:
`cos(t)=cos(t/2+t/2)`
`-7//9=cos^2(t/2)-sin^2(t/2)`
`-7/9=cos^2(t/2)-(1-cos^2(t/2))`
`-7/9=2cos^2(t/2)-1`
`-7/9+1=2cos^2(t/2)`
`2/9=2cos^2(t/2)`
`1/9=cos^2(t/2)`
`+-sqrt(1/9)=+-1/3=cos(t/2)`
`cos(t/2)=-1/3` , it has a -ve value because t/2 lie at 2nd quadrant.

B:
`sin(t)=sin(t/2+t/2)`
`-sqrt32/9=2sin(t/2)cos(t/2)`
`-sqrt32/9=2sin(t/2)(-1/3)`, from answer above `cos(t/2)=-1/3`
`-sqrt32/9=-2/3sin(t/2)`
`-sqrt32/9*-3/2=sin(t/2)`
`-sqrt(16*2)/9*-3/2=sin(t/2)`
`-4sqrt(2)/9*-3/2=sin(t/2)`
`(2sqrt2)/3=sin(t/2)`