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,

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

How do you express 2costheta + 2cos(theta + pi/3)2costheta + 2cos(theta + pi/3) in the form Rcos(theta + alpha)Rcos(theta + alpha), where R>0R>0 and 0< alpha <pi/20< alpha <pi/2?