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 $sinAcosB+cosAsinB=sin(A+B)$

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