Evaluate:
i) `sin{pi/3-sin^(-1)(-1/2)}`
ii) `sin(1/2cos^(-1)4/5)`
iii) `sin(cot^(-1)x)`
iv) `tan1/2(cos^(-1)sqrt(5)/3)`

Explanation

i) We know that `sin^(-1)(-theta)=-sin^(-1)theta`.
`therefore sin{pi/3-sin^(-1)(-1/2)}=sin{pi/3+sin^(-1)1/2}=sin(pi/3+pi/6)[therefore sin^(-1)1/2=pi/6]`
`=sin pi/2=1`.
ii) Let `cos^(-1)4/5=theta`, where `theta in [0,pi]`.
Then, `costheta=4/5`.
Since, `theta in [0,pi] rArr 1/2 theta in [0,pi/2] rArr sin1/2theta gt 0`.
`therefore sin(1/2cos^(-1)4/5)=sin1/2theta=sqrt((1-costheta)/(2)) = sqrt((1-(4/5))/(2)=1/sqrt(10)`.
ii) Let `cot^(-1)x=theta`. Then, `theta in [0,pi]`.
`therefore sin(cot^(-1)x)=sin theta gt 0`.
Now, `sin theta=1/("cosec"theta)=1/sqrt(1+cot^(2)theta)=1/sqrt(1+x^(2))`.
`therefore sin(cot^(-1)x)=sintheta=1/sqrt(1+x^(2))`.
iv) Let `cos^(-1)sqrt(5)/3=theta`. Then, `costheta=sqrt(5)/3`, where `theta in [0,pi]`.
`therefore tan1/2(cos^(-1)sqrt(5)/3)=tan1/2theta=sqrt((1-costheta)/(1+costheta))=sqrt((1-sqrt(5)//3)/(1+sqrt(5)//3))=sqrt((3-sqrt(5))/(3+sqrt(5)) xx (3-sqrt(5))/(3-sqrt(5)))=(3-sqrt(5))/(2)`.