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Answer 1 · 2018-03-02T17:18:07

d

if #sintheta# =7/13 then #theta#= #sin^-1(7/13)#

#cos(sin^-1(7/13)/2)# =0.959..... = d part 1

#cos(2sin^-1(7/13))#= #71/169# = d part 2

Answer 2 · 2018-03-02T17:24:14

The correct answer is #"option (d)"#

Apply the formula

#cos2theta=1-2sin^2theta#

#sintheta=7/13#

Therefore,

#cos2theta=1-2*(7/13)^2=1-2*49/169=71/169#

#costheta=2cos^2(theta/2)-1#

#cos^2(theta/2)=(1+costeta)/2#

#cos(theta/2)=sqrt((1+costheta)/2)#

#costheta=sqrt(1-sin^2theta)=sqrt(1-(7/13)^2)=sqrt(120/169)#

#=sqrt120/13#

Therefore,

#cos(theta/2)=sqrt((1+costheta)/2)=sqrt((1+sqrt120/13)/2)#

#=sqrt((13+2sqrt30)/26)#