# Trigonometry specific problem

This was all the information given

$$\sin^2{2 x} - \sin x-1 = 0, \ x \in [0,2\pi)$$

I did the quadratic formula and ended up with two answers which was a positive and negative. I canceled the positive one because it was higher than 1. This was my negative answer

$\sin x=1-\frac{\sqrt{5}}{2}$ . I did the inverse of that and got $x: -.6662394$.

After I got the $x$ My teacher graphed it on the $3$ and $4$ quadrant but I don't know why he graphed it on the $3$ and $4$ quadRANT.

• There are two solutions to $\sin x=\frac{1-\sqrt 5}{2}$ in $[0,2\pi)$. That's probably what your teacher tried to visualize. – Daniel R Nov 26 '14 at 14:16
• Yep I got that and then I did the inverse of that. Then I got -.662394 but after that I couldn't follow my teacher since he graphed it on the 3rd and 4th quadrant. So, I want to know why he graphed on the 3rd and 4th quadrant. – Dhondup Tenzing Nov 26 '14 at 14:18
• First, your solution is not in $[0, 2 \pi)$. Second, you will find two solutions in the 3rd and 4th quadrants respectively. – Daniel R Nov 26 '14 at 14:21

The answer you got $1-\sqrt{5/2}$ is a negative number. And the sine is negative in third and fourth quadrants.

If you don't know why is it negative , construct a circle with the center at the origin and in each quadrant draw a radius(line segment ). You'll notice that in the third and that the opposite side is negative but hypotenuse is positive because it's always greater than other two sides. And $\sin(195^{\circ}) = 1-\sqrt{5/2}$

• okay thanks so say on the question if it was cosine then I would see in which quadrant cosine is negative and positive right? – Dhondup Tenzing Nov 26 '14 at 14:23
• Remember that the hypotenuse is always positive where as other two sides may Change signs . – E and pi Nov 26 '14 at 14:25
• In case of cosine , the adjacent side in second quadrant is negitive ( because x-coordinate is negitive) and again in third quadrant(x- coordinate ) is negitive . – E and pi Nov 26 '14 at 14:27
• Thanks My buddy thought me the "All Students Take Calculus" thing so but this really helped! – Dhondup Tenzing Nov 26 '14 at 14:28
• The hypotenuse is positive because it is a side length. The lengths of the legs are positive for the same reason. Thus, the opposite side cannot be negative. The reason the sine is negative in the third and fourth quadrants is that the sine is defined to be the $y$-coordinate of the point where the terminal side of the angle intersects the unit circle. – N. F. Taussig Nov 26 '14 at 15:42

If $\theta$ is an angle in standard position (the vertex is at the origin and the initial side lies on the positive $x$-axis), $\sin\theta$ is defined to be the $y$-coordinate of the point where the terminal side of the angle intersects the unit circle. Thus, if $\sin\theta < 0$, the terminal side of the angle must lie in the third quadrant, on the negative $y$-axis, or in the fourth-quadrant. Since

$$\sin x = \frac{1 - \sqrt{5}}{2}$$

and

$$-1 < \frac{1 - \sqrt{5}}{2} < 0$$

the terminal side of angle $x$ must lie in the third quadrant or fourth quadrant.