Question #8ab7c

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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!