$f(\theta) = \cos \theta - \sqrt{3}\sin \theta$ in the form $r\sin(\theta-\alpha)$

Question

My answer for this question was $2\sin(θ - 30)$, but the answer in the textbook says $-2\sin(θ - 30)$. I tried to use the concept of odd functions so using the idea that $-\sin(θ)= \sin(-θ)$.

So $-2\sin(θ-30) = 2\sin(30-θ)$ but that still is not the same as my answer and I was wondering why?

In addition, the question also asks to give the maximum and minimum values of the function and the values between $0$ and $360$ at which they occur.

I get:

max value $y = 2$ max $x$ at $θ = 120$

min $y$ value = $ -2$ min $x$ at $θ = 300$

But the textbook answers say:

max value $y = 2$ at $θ = 300$ min value $y = -2$ at $θ = 106.3$

I am quite confused as to why.

Thank you.

Answer 1

Hint: Assume that the expression can be rewritten as

$$A\sin(\theta+\phi)=A\sin\theta\cos\phi + A\cos\theta \sin \phi$$

I used the formula for the sum of angles for the sine function. Now, compare your expression to this expression to obtain

$$-\sqrt{3}=A\cos \phi$$ $$1=A\sin \phi.$$

Square both equations and add them to obtain $$A^2\left[\sin^2\phi + \cos^2\phi \right]=1^2+(-\sqrt{3})^2$$ $$\implies A^2 = 1+3$$ $$\implies A^2=4.$$ I used the theorem of Pythagoras. Now, divide both equations to obtain $$\dfrac{\sin \phi}{\cos \phi}= \dfrac{1}{-\sqrt{3}}.$$ $$\implies \tan \phi = -1/\sqrt{3}.$$ Can you determine $A$ and $\phi$ from that? You can use $\phi=-\alpha$ to obtain your value for alpha.

Answer 2

\begin{align}\cos\theta-\sqrt3\sin\theta&=2\left(\frac12\cos\theta-\frac{\sqrt3}2\sin\theta\right)\\&=-2\left(\sin(-30^\circ)\cos\theta+\cos(-30^\circ)\sin\theta\right)\\&=-2\sin(\theta-30^\circ).\end{align}Therefore, your textbook is right.

Answer 3

Hint: Write your function in the form $$f(\theta)=4\left(\frac{1}{4}\cos(x)-\frac{\sqrt{3}}{4}\sin(x)\right)$$