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

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 $$θ = 300$$

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

I am quite confused as to why.

Thank you.

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.

• Ah! I thought that 1 becomes -1, which is where i made my error. Why is it that you do 1 and why is it that you use this 𝐴sin(𝜃+𝜙) expression instead of the one with minus. The question requries me to use 𝐴sin(𝜃-𝜙). Commented Apr 27, 2019 at 14:43
• You can replace $\phi = -\alpha$ to obtain your form. I don't like to put a minus in there when it is not necessary. What do you mean by "Why is it that you do $1$"? Commented Apr 27, 2019 at 14:46
• The question for why is it that you do 1, was answer by telling me to replace 𝜙=−𝛼, so thank you for that! I was also wondering why the minus value for theta is 106.3? Do you think it was an error on the textbook? Commented Apr 27, 2019 at 15:17
• $\tan \phi=-1/\sqrt{3}$ has two solutions in the unit circle $\phi_1 = -\pi/6$ and $\phi_2 = 5/6\pi$. By looking at the signs of $\cos \phi=-\sqrt{3}/2$ and $\sin \phi = 1/2$ we see that $\phi_2$ is the correct solution. Then $2\sin(x-5/6\pi)$ can be rewritten to the form in your textbook. Commented Apr 27, 2019 at 15:31

\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.

• Sir in the RHS of the 1st equation, the coeff. of sin(theta) is sqrt(3)/2, please change it. Commented Apr 27, 2019 at 14:38
• I've edited my answer. Thank you. Commented Apr 27, 2019 at 14:39
• Do you know why i was wrong for the max/min x values. Because Im not too sure where theta = 106.3 comes from? Commented Apr 27, 2019 at 14:42

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