How do you rewrite #sqrt(3)sin theta + costheta# as a sum?

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Answer 1 · 2016-11-23T14:29:35

We know that #sin^2theta + cos^2theta = 1#. Isolating #cos^2theta#, we have:

#cos^2theta = 1 - sin^2theta#

#costheta = sqrt(1 - sin^2theta)#

#=>sqrt(3)sin theta + sqrt(1 - sin^2theta)#

Hopefully this helps!

Answer 2 · 2016-11-23T16:03:05

Given expression

#=sqrt3sintheta+costheta#

#=>2(sqrt3/2sintheta+1/2costheta)#

#=>2(cos(pi/6)sintheta+sin(pi/6)costheta)#

#=>2(sinthetacos(pi/6)+costhetasin(pi/6))#

#=>2sin(theta+pi/6)#
(applying formula #sinAcosB+cosAsinB=sin(A+B)#