Question

How do you find the exact value of, $\sin \left( \dfrac{19 \pi}{12} \right)$ ?

Answer 1

Hint: To find the exact value of $\sin \left( \dfrac{19 \pi}{12} \right)$, we will first of all find the quadrant of this angle. As one can see, this angle is greater than $\left( \dfrac{18 \pi}{12} \right)$ and less than $\left( \dfrac{24pi}{12} \right)$, then it means the angle lies in the fourth quadrant. The “sine” of any angle in the fourth quadrant can be calculated using the formula, $\left[ \sin \theta =-\sin \left( {{360}^{\circ }}-\theta \right):{{270}^{\circ }}\le \theta \le {{360}^{\circ }} \right]$ . We shall proceed using this formula to get our solution.

Complete step-by-step solution:
The first step towards our solution will be to convert the angle in radians to angle in degrees as this makes the problem easy to visualize and solve. This can be done by putting the value of 1 pi radian as ${{180}^{\circ }}$. On doing so we get the new expression as:
$\begin{align}
  & =\sin \left( \dfrac{19 \pi}{12} \right) \\
 & =\sin \left( \dfrac{19\times {{180}^{\circ }}}{12} \right) \\
 & =\sin \left( {{285}^{\circ }} \right) \\
\end{align}$
Now, this angle in fourth quadrant could be calculated as follows:
$\begin{align}
  & \Rightarrow \sin \left( {{285}^{\circ }} \right)=-\sin \left( {{360}^{\circ }}-{{285}^{\circ }} \right) \\
 & \Rightarrow \sin \left( {{285}^{\circ }} \right)=-\sin \left( {{75}^{\circ }} \right) \\
\end{align}$
Now, the angle operated by ‘sine’ could be split into two standard angles whose values are known. And then, we could apply the sum of angles formula for a sine function which is given as:
$\Rightarrow \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
Applying these steps simultaneously, we get:
$\Rightarrow \sin \left( {{285}^{\circ }} \right)=-\sin \left( {{30}^{\circ }}+{{45}^{\circ }} \right)$
$\begin{align}
  & \Rightarrow \sin \left( {{285}^{\circ }} \right)=-\left[ \sin {{30}^{\circ }}\cos {{45}^{\circ }}+\cos {{30}^{\circ }}\sin {{45}^{\circ }} \right] \\
 & \Rightarrow \sin \left( {{285}^{\circ }} \right)=-\left[ \dfrac{1}{2}\times \dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2}\times \dfrac{1}{\sqrt{2}} \right] \\
 & \Rightarrow \sin \left( {{285}^{\circ }} \right)=-\left[ \dfrac{1}{2\sqrt{2}}+\dfrac{\sqrt{3}}{2\sqrt{2}} \right] \\
 & \Rightarrow \sin \left( {{285}^{\circ }} \right)=-\left[ \dfrac{\sqrt{3}+1}{2\sqrt{2}} \right] \\
\end{align}$

Using the value of:
$\sqrt{3}\text{ as 1}\text{.732}$, and
$\sqrt{2}\text{ as 1}\text{.414}$
We get the final result of our solution as:
$\begin{align}
  & \Rightarrow \sin \left( {{285}^{\circ }} \right)=-\left[ \dfrac{1.732+1}{2\times 1.414} \right] \\
 & \Rightarrow \sin \left( {{285}^{\circ }} \right)=-\left[ \dfrac{2.732}{2.828} \right] \\
 & \therefore \sin \left( {{285}^{\circ }} \right)=-0.966 \\
\end{align}$
Hence, the exact value of, $\sin \left( \dfrac{19 \pi}{12} \right)$ comes out to be $-0.966$.

Note: In lengthy trigonometric calculations like these, one should be very careful of concurring any silly mistakes in the solution. Also, we should know the square root values of very common numbers like square root of 2, square root of 3, square root of 5 and so on up to square root of 8. Their values may or may not be specified in the problem, so one should remember these in order to avoid any difficulties while solving such problems.