Question

How do I find the value of $\sin 330$?

Answer 1

Hint:The value of the sine function can be found by rewriting the given function that is $\sin 330$ as $\sin (360 - 30)$ . The function $\sin (360 - 30)$ is in the form of $\sin (A - B)$ which is given by: $\sin (A - B) = \sin A.\cos B - \cos A.\sin B$ so now by substituting $A = 360$ and $B = 30$ and simplifying the expression we get required answer.

Complete step-by-step answer:
The given function is $\sin 330$ for which we do not have the direct values in the standard trigonometric ratio table.
Now we need to rewrite the function $\sin 330$ as $\sin (360 - 30)$ , which makes simplification easier. The function $\sin (360 - 30)$ is in the form of $\sin (A - B)$. The $\sin (A - B)$ formula is given by:$\sin (A - B) = \sin A.\cos B - \cos A.\sin B$ . Now we can use this formula to find the required answer.
Therefore, now by substituting $A = 360$ and $B = 30$ in the above given formula, we get
$\sin (360 - 30) = \sin 360.\cos 30 - \cos 360.\sin 30$
We know that the values for the above functions which is given by:
 $\
  \sin 30 = \dfrac{1}{2} \\
  \sin 360 = 0 \\
  \cos 30 = \dfrac{{\sqrt 3 }}{2} \\
  \cos 360 = 1 \\
\ $
Now substitute these values in the above expression for simplification purpose, we get
$ \Rightarrow \sin (360 - 30) = \left( {0 \times \dfrac{{\sqrt 3 }}{2}} \right) - \left( {1 \times \dfrac{1}{2}} \right)$
On simplifying the above expression, we get
$ \Rightarrow \sin (360 - 30) = - \dfrac{1}{2}$

Additional information:
The below table gives the details of the values of the trigonometric functions:
\[\begin{array}{*{20}{c}}
  {}&0&{\dfrac{\pi }{6}}&{\dfrac{\pi }{4}}&{\dfrac{\pi }{3}}&{\dfrac{\pi }{2}} \\
  {\sin \theta }&0&{\dfrac{1}{2}}&{\dfrac{1}{{\sqrt 2 }}}&{\dfrac{{\sqrt 3 }}{2}}&1 \\
  {\cos \theta }&1&{\dfrac{{\sqrt 3 }}{2}}&{\dfrac{1}{{\sqrt 2 }}}&{\dfrac{1}{2}}&0 \\
  {\tan \theta }&0&{\dfrac{1}{{\sqrt 3 }}}&1&{\sqrt 3 }&\infty \\
  {\csc \theta }&\infty &2&{\sqrt 2 }&{\dfrac{2}{{\sqrt 3 }}}&0 \\
  {\sec \theta }&1&{\dfrac{2}{{\sqrt 3 }}}&{\sqrt 2 }&2&\infty \\
  {\cot \theta }&\infty &{\sqrt 3 }&1&{\dfrac{1}{{\sqrt 3 }}}&0
\end{array}\]

Note: The given problem can be solved in another way that is by using quadrants. The given function is $\sin 330$ where the angle 330 lies in between 270 degrees and 360 degrees which means the given function lies in the fourth quadrant where the sine function is negative. Therefore we can write as $\sin 330 = \sin (360 - 30)$ here we know that $\sin (360 - \theta ) = - \sin \theta $ . Hence we can write as $\sin 330 = - \sin 30 = - \dfrac{1}{2}$ . If we do not remember the formula then we can use this process, or we can use a calculator to find this.