Taking `theta=30^(@)`, verify each of the following
`(i) sin 2theta=2sin theta cos theta`
`(ii) cos theta=2cos^(2)theta-1=1-2sin^(2)theta`
`(iii) tan2theta=(2tantheta)/(1-tan^(2)theta)`

Explanation

When `theta=30^(@)`, we have `2theta=60^(@)`.
`(i)sin 2theta=sin60^(@)=(sqrt(3))/(2)`
`2sinthetacostheta=2sin30^(@)cos30^(@)=(2xx(1)/(2)xx(sqrt(3))/(2))=(sqrt(3))/(2)`.
`:.sin2theta=2sinthetacostheta`.
`(ii) cos2theta=cos60^(@)=(1)/(2)`.
`2cos^(2)theta-1=2xxcos^(2)30^(@)-1`
`=[2xx((sqrt(3))/(2))^(2)-1]=(2xx(3)/(4)-1)=((3)/(2)-1)=((3)/(2)-1)=(1)/(2)`.
`1-2sin^(2)theta=1-2sin^(2)30^(@)`
`=[1-2xx((1)/(2))^(2)]=(1-2xx(1)/(4))=(1-(1)/(2))=(1)/(2)`.
`:.cos2theta=2cos^(2)theta-1=1-2sin^(2)theta`.
`tantheta=tan60^(@)=sqrt(3)`
`(2tantheta)/(1-tan^(2)theta)=(2tan30^(@))/(1-tan^(2)30^(@))=(2xx(1)/(sqrt(3)))/(1-((1)/(sqrt(3)))^(2))=(((2)/(sqrt(3))))/((1-(1)/(3)))`
`=((2)/(sqrt(3))xx(3)/(2))=sqrt(3)`.
`:.tan2theta=(2tantheta)/(1-tan^(2)theta)`