Question

What is the value of ${{180}^{\circ }}$?
A.1
B.0
C.-1
D.$\dfrac{1}{2}$

Answer 1

Hint: Convert the given obtuse angle inside the sine function to acute angle from $\left( {{0}^{\circ }}-{{90}^{\circ }} \right)$ with the help of the relation $\sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta $. Value of $\sin \theta $ is 0.

Complete step-by-step answer:
As we need to calculate the value of $\sin {{180}^{\circ }}$. So, as we know the values of trigonometric functions at some certain angles only that are ${{0}^{\circ }},{{45}^{\circ }},{{30}^{\circ }},{{60}^{\circ }},{{90}^{\circ }}$. It means we don’t have any direct value of $\sin {{180}^{\circ }}$. So, we need to convert the angle ${{180}^{\circ }}$ to any acute angle i.e. between ${{0}^{\circ }}-{{90}^{\circ }}$, and hence, we can get the value of it as we know values of trigonometric functions only for some certain angles as discussed above.
Now, we can use the property of trigonometric functions related for converting obtuse angle to acute angle. So, as we know the relation of trigonometric functions as
$\sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta \ldots \left( i \right)$
Equation $\left( i \right)$ is the proved relation and it can be done with the help of property of trigonometric functions in the quadrants. So, as we need to find value of _{$\sin {{180}^{\circ }}$}, so put $\theta ={{0}^{\circ }}$ to the equation $\left( i \right)$and hence, we get $\sin \left( {{180}^{\circ }}-{{0}^{\circ }} \right)=\sin {{0}^{\circ }}$
$\sin {{180}^{\circ }}=\sin {{0}^{\circ }}\ldots \left( ii \right)$
As, we know the value of $\sin {{0}^{\circ }}$ is 0. So, the value of $\sin {{180}^{\circ }}$ will be ‘0’ as well from the equation$\left( ii \right)$.
Hence, the value of $\sin {{180}^{\circ }}$ is given as 0.
Note: Another approach for calculating value of $\sin {{180}^{\circ }}$ would be that we can write${{180}^{\circ }}={{90}^{\circ }}+{{90}^{\circ }}$. And use identity as $sin\left( {{90}^{\circ }}+\theta \right)=\cos \theta $
Put $\theta ={{90}^{\circ }}$, we get
$sin\left( {{180}^{\circ }} \right)=\cos ({{90}^{o}})=0$
Always try to convert obtuse angles involved with any trigonometric functions to acute angle form. As we know, the values of trigonometric functions are only $\left( 0-{{90}^{\circ }} \right)$.