Why the complex number $6(\sin(240^{\circ}) + i \cos(240^{\circ}))$ lies in first quadrant?

Question

problem :

find the quadrant in which $6(\sin(240^{\circ}) + i \cos(240^{\circ}))$ lies and find the principal argument then rewrite the complex number

my attempt :

$6(\sin(240^{\circ}) + i \cos(240^{\circ})) = -3\sqrt{3} -3i$ then it should lie in the $3^{rd}$ quad

$\theta = \tan^{-1}({-3 \over -3\sqrt{3} }) = 30^\circ$

the postive angle indicates that it's measured in the anti-clockwise direction

the principle argument = $-180^{\circ} + \theta = -180^{\circ} + 30^{\circ} = -150^\circ$

then the complex number = $6(\cos(-150^{\circ}) + i\sin(-150^{\circ}))$

and it should look like this on plane :

however my textbook had a totally different opinion

it said that the complex number $6(\sin(240^{\circ}) + i \cos(240))$ lies in first quadrant you can take a look at the textbook's answer yourself :

We had the same answer at the end but the textbook used a different method which I did not understand and I am very confused right now

Any help will be appreciated

Thanks in advance.

Answer 1

First off, yeah, that first quadrant stuff was totally wrong.

The solution method that they used did not calculate the trig functions as you did (essentially transforming the number from polar form into rectangular). Instead, they tried to convert the number directly into $r(\cos\theta+i\sin\theta)$ form from its original form where the sine and cosine were reversed. For that, they used the trig identities that $\sin\theta=\cos(\frac\pi2-\theta)$ and $\cos\theta=\sin(\frac\pi2-\theta)$.

Answer 2

It's an error from the textbook. The authors are right when they assert that the given number is equal to$$6\left(\cos\left(-\frac{5\pi}6\right)+\sin\left(-\frac{5\pi}6\right)i\right),$$but the conclusion should be that it belongs to the third quadrant. This also follows from the fact that $\sin\left(\frac{4\pi}3\right),\cos\left(\frac{4\pi}3\right)<0$.

Answer 3

I think this is a terrible "solution" in the textbook. It takes an easy matter and unnecessarily complicates it with a lot of nonsense. The angle of $\theta=240^{\circ}=\frac{4\pi}{3}$ lies in quadrant III, and the value of $r=6$ is positive, so no flipping over the origin and the point stays in quadrant III — end of story. And your original calculation $6(\sin(240^{\circ})+i\cos(240^{\circ}))=-3\sqrt{3}-3i$ and conclusion were perfectly correct.