Question

Rewrite ℎ(x) = √3 sin(x) + cos(x) in the form ℎ(x) = a cos (x − c).

Answer 1

`\cos(x-0.523) = (1)/(2)`

Explanation:

We are given the following in the question:

`h(x) = √(3) \sin(x) + \cos(x)`

We have to write this in

`h(x) = R\cos (x-c)`

We can do this in the following manner:

`a = R\cos c\\b = R\sin c\\\text{Putting vallues we get}\\h(x) = a\sin x + b\cos x\\= R\cos c \sin x + R \sin c \cos x\\=R\cos(x-c)\\R = √(a^2 + b^2)\\\\\tan c = \displaystyle(b)/(a)`

Thus, we can write:

`R = \sqrt{√(3)^2 + 1^2} = √(3+1) = 2\\h(x) = R\cos(x-c)\\h(x) = 2\cos(x-c)\\\\\tan c = (1)/(√(3))\\\\c = \tan^(-1)(1)/(√(3)) = 0.523\text{ radians}\\\\\cos(x-0.523) = (1)/(2)`