How do you express #2costheta + 2cos(theta + pi/3)# in the form #Rcos(theta + alpha)#, where #R>0# and #0< alpha <pi/2#?

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Answer 1 · 2018-05-20T09:19:19

#R=2sqrt3 and alpha=pi/6 in (0,pi/2)#.

Let, #f(theta)=2costheta+2cos(theta+pi/3)#.

Using #cosx+cosy=2cos((x+y)/2)cos((x-y)/2)#, we have,

#f(theta)=2{2cos((theta+(theta+pi/3))/2)cos((theta-(theta+pi/3))/2)}#,

#=4cos(theta+pi/6)cos(-pi/6)#,

#=4cos(theta+pi/6)*(sqrt3/2)#.

# rArr f(theta)=2sqrt3cos(theta+pi/6)#.

So, we have, #f(theta)=Rcos(theta+alpha); R gt 0, theta in (0,pi/2)#, where,

#R=2sqrt3 and alpha=pi/6 in (0,pi/2)#.