How do you find the value for #sin2theta#, #cos2theta#, and #tan2theta# and the quadrant in which #2theta# lies given #sintheta=4/5# and #theta# is in quadrant I?

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Answer 1 · 2016-10-01T16:17:41

#sin2theta=24/25#, #cos2theta=-7/25# and #tan2theta=-24/7#

As #sintheta=4/5# and it is first quadrant, all trigonometric ratios of #theta# are positive.

and #costheta=sqrt(1-(4/5)^2)#

= #sqrt(1-16/25)=sqrt(9/25)=3/5#

and #tantheta=(4/5)/(3/5)=4/5xx5/3=4/3#

Now #sin2theta=2sinthetacostheta=2xx4/5xx3/5=24/25#

#cos2theta=2cos^2theta-1=2xx(3/5)^2-1#

= #2xx9/25-1=18/25-1=-7/25#

#tan2theta=(24/25)/(-7/25)=24/25xx25/(-7)=-24/7#