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Answer 1 · 2018-01-05T05:04:22

# cos(x-5pi/6)=1/2{sqrt(1-cos^2x)-sqrt3cosx}#.

Prerequisites #(1): cos(A-B)=cosAcosB+sinAsinB#,

#(2)(i): cos(pi-theta)=-costheta, (2)(ii) : sin(pi-theta)=sintheta#,

#(3)(i): cos(pi/6)=sqrt3/2, (3)(ii) sin(pi/6)=1/2#,

#(4): sintheta=sqrt(1-cos^2theta)#.

Now, #cos(x-5pi/6)=cosxcos(5pi/6)+sinxsin(5pi/6)#,

#=cosxcos(pi-pi/6)+sinxsin(pi-pi/6)#,

#=(-cos(pi/6))cosx+sin(pi/6)sinx#,

#=-sqrt3/2cosx+1/2sinx#,

#=1/2(sinx-sqrt3cosx)#.

#rArr cos(x-5pi/6)=1/2{sqrt(1-cos^2x)-sqrt3cosx},# is the

desired expression!