If `cosA=24/25`, evaluate the following, given that `0 lt A lt pi/2`.
i) sin2A
ii) cos 2A
iii) tan 2A

Explanation

`cosA=24/25` where A lies in 1st quadrant.
`therefore sin^(2)A=1-cos^(2)A=1-(24/25)^(2)=49/625`
`rArr sinA=7/25` (A lies in 1st quadrant)
and `tanA= (sinA)/(cosA)=(7/25)/(24/25)=7/24`
Now (i) `sin2A=2sinAcosA=2.7/25.24/25=336/625` Ans.
ii) `cos2A=2cos^(2)A-1`
`2.(24/25)^(2)-1`
`=(1152-625)/(625)=527/625`. Ans.
iii) `tan2A=(2tanA)/(1-tan^(2)A)=(2(7/24))/(1-(7/24))^(2)`
`=(2 xx 7 xx 24)(576-49)=336/527` Ans.