Question

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

Answer 1

Hint: In this question, we will use the formula of trigonometric ratios for complementary and supplementary angles and the all sin tan cos sign convention to find values of given expressions.

Step-by-step answer:
In trigonometry, if we know the value of one of trigonometric functions for some angle, then we can find the value of that trigonometric function for angle ${{90}^{\circ }}$ more than that angle and also for ${{90}^{\circ }}$ less than that angel by one formula and a rule. This formula for sine function, is given as,
$\sin \left( {{90}^{\circ }}n\pm x \right)=\left\{ \begin{matrix}
   \pm \cos x & \text{if}\,n\,\text{is}\,\text{odd} \\
   \pm \sin x & \text{if}\,n\,\text{is}\,\text{even}\, \\
\end{matrix} \right.\cdots \cdots \left( i \right)$,
Where, $n$ is a natural number.
Now, sign in front of the trigonometric function, in this case sine and cosine, will depend on the quadrant in which the angle ${{90}^{\circ }}n\pm x$ will lie. If the angle lies in the first quadrant, the sign will be positive for all the trigonometric functions. If the angle lies in the second quadrant, only values of sine function and its reciprocal function, that is, cosecant will be positive. If angle lies in the third quadrant, then only tangent function and its reciprocal function cotangent function will be positive. In the third quadrant, only value cosine function and its reciprocal function, secant function will be positive.
Also, sine function is an odd function. That is,
$\sin (-x)=-\sin x\cdots \cdots \left( ii \right)$
Now, we can write,
$\begin{align}
  & \sin {{60}^{\circ }}=\sin \left( {{90}^{\circ }}-{{30}^{\circ }} \right) \\
 & =\sin \left( {{90}^{\circ }}\times 1-{{30}^{\circ }} \right) \\
\end{align}$
Here, $n$ is 1, which is odd. And, ${{60}^{\circ }}$ lies in the first quadrant, so the sign will be positive.
Therefore, using equation $\left( i \right)$, we get,
$\sin {{60}^{\circ }}=\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$.
Again, we have,
$\begin{align}
  & \sin {{120}^{\circ }}=\sin \left( {{90}^{\circ }}+{{30}^{\circ }} \right) \\
 & =\sin \left( {{90}^{\circ }}\times 1+{{30}^{\circ }} \right) \\
\end{align}$
Here, $n$ is 1, which is odd. And, ${{120}^{\circ }}$ lies in the second quadrant, so sign will be positive.
Therefore, using equation $\left( i \right)$, we get,
$\sin {{120}^{\circ }}=\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$.
Again, we can write,
$\begin{align}
  & \sin {{240}^{\circ }}=\sin \left( {{270}^{\circ }}-{{30}^{\circ }} \right) \\
 & =\sin \left( {{90}^{\circ }}\times 3-{{30}^{\circ }} \right) \\
\end{align}$
Here, $n$ is 3, which is odd. And, ${{240}^{\circ }}$ lies in the third quadrant, so the sign will be negative.
Therefore, using equation $\left( i \right)$, we get,
$\sin {{240}^{\circ }}=-\cos {{30}^{\circ }}=-\dfrac{\sqrt{3}}{2}$.
Again, we can write,
$\begin{align}
  & \sin {{300}^{\circ }}=\sin \left( {{270}^{\circ }}+{{30}^{\circ }} \right) \\
 & =\sin \left( {{90}^{\circ }}\times 3+{{30}^{\circ }} \right) \\
\end{align}$
Here, $n$ is 3, which is odd. And, ${{300}^{\circ }}$ lies in the fourth quadrant, so the sign will be negative.
Therefore, using equation $\left( i \right)$, we get,
$\sin {{300}^{\circ }}=-\cos {{30}^{\circ }}=-\dfrac{\sqrt{3}}{2}$.
Lastly, using equation $\left( ii \right)$, we can write,
$\sin \left( -{{30}^{\circ }} \right)=-\sin {{30}^{\circ }}=-\dfrac{1}{2}$.
Hence, we got the values of sine for all given angles.

Note: In this type of question, for placing a sign, we need to check the quadrant for the given trigonometric function and not the one which we get from the formula, that is, in front of which sign is to be placed.