Question

Given that $\sin {30}^{\circ }=1/2$ and $\cos {30}^{\circ }=\sqrt{3}/2$. Determine the values of $\sin {60}^{\circ },\sin {120}^{\circ },\sin {240}^{\circ },\sin {300}^{\circ }$, and $\sin (-{30}^{\circ })$.

Answer 1

1. First, we should determine the quadrant in which ${60}^{\circ }$ lies.
It is obviously the first quadrant. Then, we should recall whether sine in the first quadrant is positive or negative. "All Silver Tea Cups" tells us that all the trigonometric ratios are positive in the first quadrant; therefore, $\sin {60}^{\circ }$ must be positive.
Now, we should write ${60}^{\circ }$ in such a way that it is $\pm {30}^{\circ }$ with any of the two axes (the horizontal $XO{X}^{\prime }$ and the vertical $YO{Y}^{\prime }$) . So we can write $\sin {60}^{\circ }=\sin ({90}^{\circ }-{30}^{\circ })$. Now, we can recall from the trigonometric ratios in the previous page that $\sin ({90}^{\circ }-\theta )=\cos \theta$
$\Rightarrow \sin {60}^{\circ }=\sin ({90}^{\circ }-{30}^{\circ })=\cos {30}^{\circ }=\sqrt{3}/2$.
2. Similarly, we can find out the value of $\sin {120}^{\circ }$. This angle lies in the second quadrant. In the second quadrant, sine is positive. Therefore, $\sin {120}^{\circ }=$ some positive value.
$\sin {120}^{\circ }=\sin ({90}^{\circ }+{30}^{\circ })=\cos {30}^{\circ }=\sqrt{3}/2$
3. $\sin {240}^{\circ }$ lies in the third quadrant. So it should be negative.
$\sin {240}^{\circ }=\sin ({270}^{\circ }-{30}^{\circ })=-\cos {30}^{\circ }=-\sqrt{3}/2$
4. $\sin {300}^{\circ }$ lies in the fourth quadrant. So it should be negative.
$\sin {300}^{\circ }=\sin ({270}^{\circ }-{30}^{\circ })=-\cos {30}^{\circ }=-\sqrt{3}/2$
5. $\sin (-{30}^{\circ })$ lies in the fourth quadrant. So it should be negative.
$\sin (-{30}^{\circ })=-\sin {30}^{\circ }=-1/2$.