If #cos2theta=2/3# and #theta# lies in #Q2#, find #sintheta# and #costheta#?

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Answer 1 · 2017-11-21T05:59:21

#sintheta=sqrt6/6# and #costheta=-sqrt30/6#

#sintheta=sqrt((1-cos2theta)/2)#

= #sqrt((1-2/3)/2)#

= #sqrt(1/6)#

= #1/sqrt6#

= #sqrt6/6#

#costheta=-sqrt((1+cos2theta)/2)#

= #-sqrt((1+2/3)/2)#

= #-sqrt(5/6)#

= #-sqrt5/sqrt6#

= #-sqrt30/6#