If #cosx=2/5#, how do you find #sin2x#?

Question

10100 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-30T22:34:45

#sin 2x=2/5 sqrt(3/5)#

#cos x=2/5" ; "sin 2x=?#

#sin2x=sin x*cos x+cos x*sin x#

#sin 2x=2*sin x*cos x#

#"so ;" sin x=sqrt(1-cos^2 x)#

#"we can write as ;"#

#sin 2x=sqrt (1-cos^2 x)*cos x#

#sin 2x=sqrt(1-2/5)*2/5#

#sin 2x=sqrt((5-2)/5)*2/5#

#sin 2x=sqrt(3/5)*2/5#

#sin 2x=2/5 sqrt(3/5)#