Question

Question #49b10

Answer 1

`2sqrt(3)cos(theta + pi/3)`

Explanation:

The general form is `a*cos(theta)-b*sin(theta) = R*cos(theta+alpha)`, where `R = sqrt(a^2+b^2)` and `alpha=tan^-1(b/a)`.

First calculate `R = sqrt(a^2+b^2)= sqrt((sqrt(3))^2+(3)^2)=sqrt(12)=2sqrt(3)`.

Next calculate `alpha=tan^-1(b/a) = tan^-1(3/sqrt(3))=tan^(-1)(sqrt(3)) = pi/3`.

So `sqrt(3)cos(theta)-3sin(theta) = 2sqrt(3)cos(theta + pi/3)`

See this video for additional explanation.