Find sin2x, cos2x, and tan2x if sinx=-3/5 and x terminates in quadrant IV?

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Answer 1 · 2018-04-25T17:28:02

#sin2x=-24/25 , cos2x =7/25 and tan2x= -24/7#

Here,

#sinx=-3/5# < 0

#x# terminates in #IV^(th) Quadrant=>cosx > 0#

#:.cosx=+sqrt(1-sin^2x)=sqrt(1-9/25)=sqrt(16/25)=4/5#

So,

#color(red)((i)sin2x=2sinxcosx)=2(-3/5)(4/5)=-24/25#

#color(blue)((ii)cos2x=cos^2x-sin^2x)=16/25-9/25=7/25#

#color(violet)((iii)tan2x=(sin2x)/(cos2x))=(-24/25)/(7/25)=-24/7#

Note:

#IV^(th) Quadrant=>(3pi)/2 < x < 2pi=>3pi < 2x <4pi#

#i.e.III^(rd) or IV^(th) Quadrant#.

But, #sin2x <0,cos2x >0 and tan2x < 0=>IV^(th)Quadrant#