Question

How do I find the value of $\sin \left( { - 30^\circ } \right)$?

Answer 1

Hint: We can expand $\sin \left( { - 30^\circ } \right)$ as $\sin \left( {30^\circ - 60^\circ } \right)$. Then we can simplify it using the identity, $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$. Then we can substitute this in the given expression. We can then give the values for trigonometric functions $\sin 30^\circ ,\cos 30^\circ ,\sin 60^\circ ,\cos 60^\circ $. After further simplification, we will get the required solution. Then we can check the options and find the correct option.

Complete step-by-step solution:
Trigonometric values of ratios like sin, cos, tan, cosec, cot, and secant are very useful while solving and dealing with problems related to the measurement of length and angles of a right-angled triangle. 0°, 30°, 45°, 60°, and 90° are the commonly used values of the trigonometric function to solve trigonometric problems.
We need to find the value of $\sin \left( { - 30^\circ } \right)$.
We can write -30 as -30 = 30 – 60. So, $\sin \left( { - 30^\circ } \right)$ will become,
$ \Rightarrow \sin \left( { - 30^\circ } \right) = \sin \left( {30^\circ - 60^\circ } \right)$
We know that,
$\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$
On applying this identity, we get
$ \Rightarrow \sin \left( { - 30^\circ } \right) = \sin 30^\circ \cos 60^\circ - \cos 30^\circ \sin 60^\circ $
We know that $\sin 30^\circ = \dfrac{1}{2},\cos 30^\circ = \dfrac{{\sqrt 3 }}{2},\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}$ and $\cos 60^\circ = \dfrac{1}{2}$. On substituting these in the above equation, we get
$ \Rightarrow \sin \left( { - 30^\circ } \right) = \dfrac{1}{2} \times \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2} \times \dfrac{{\sqrt 3 }}{2}$
Simplify the terms,
$ \Rightarrow \sin \left( { - 30^\circ } \right) = \dfrac{1}{4} - \dfrac{3}{4}$
As the denominators are equal, we can subtract the numerators.
$ \Rightarrow \sin \left( { - 30^\circ } \right) = - \dfrac{2}{4}$
Cancel out the common factors,
$ \Rightarrow \sin \left( { - 30^\circ } \right) = - \dfrac{1}{2}$

Hence, the value of $\sin \left( { - 30^\circ } \right)$ is $ - \dfrac{1}{2}$.

Note: Alternate method to solve this problem is given by,
We need to find the value of $\sin \left( { - 30^\circ } \right)$.
We know that,
$\sin \left( { - \theta } \right) = - \sin \theta $
On applying this identity, we get
$\sin \left( { - 30^\circ } \right) = - \sin 30^\circ $
We know that $\sin 30^\circ = \dfrac{1}{2}$. On substituting these in the above equation, we get
$\sin \left( { - 30^\circ } \right) = - \dfrac{1}{2}$
Hence, the value of $\sin \left( { - 30^\circ } \right)$ is $ - \dfrac{1}{2}$.
We must take care of the order while expanding the trigonometric function of the difference of the angle. We must know the values of the trigonometric function at basic angles such as 0, 30, 45, 60, and 90 degrees.