
How do you find the value of: $\sin \left( { - \dfrac{\pi }{2}} \right)$?
Answer
513.6k+ views
Hint: In the given problem, we are required to find the sine of a given angle using some simple and basic trigonometric compound angle formulae and trigonometric identities. Such questions require basic knowledge of compound angle formulae and their applications in this type of questions. We should know the periodicity of the trigonometric functions.
Complete step by step answer:
So, we have to calculate the value of $\sin \left( { - \dfrac{\pi }{2}} \right)$. Now, we know that trigonometric functions are periodic functions. Since cosine and sine function are periodic functions with periods of $2\pi $, so the value of cosine and sine gets repeated after intervals in multiples of $2\pi $. Hence, we can add or subtract the term $\left( {2\pi } \right)$ in the angle of the trigonometric function according to our choice.
\[ \Rightarrow \sin \left( { - \dfrac{\pi }{2}} \right) = \sin \left( { - \dfrac{\pi }{2} + 2\pi } \right)\]
Now, simplifying the angle of the trigonometric function, we get,
\[ \Rightarrow \sin \left( { - \dfrac{\pi }{2}} \right) = \sin \left( {\dfrac{{3\pi }}{2}} \right)\]
So, splitting up the angle \[\left( {\dfrac{{3\pi }}{2}} \right)\] as \[\left( {\pi + \dfrac{\pi }{2}} \right)\], we get,
\[ \Rightarrow \sin \left( { - \dfrac{\pi }{2}} \right) = \sin \left( {\pi + \dfrac{\pi }{2}} \right)\]
Now, we also the trigonometric formula $\sin \left( {\pi + \theta } \right) = - \sin \theta $.
\[ \Rightarrow \sin \left( { - \dfrac{\pi }{2}} \right) = - \sin \left( {\dfrac{\pi }{2}} \right)\]
We know the value of \[\sin \left( {\dfrac{\pi }{2}} \right)\] is $\left( 1 \right)$. Substituting the same, we get,
$ \therefore $$ - 1$
Hence, the value of $\sin \left( { - \dfrac{\pi }{2}} \right)$ is $\left( { - 1} \right)$.
Note: Periodic Function is a function that repeats its value after a certain interval. For a real number $T > 0$, $f\left( {x + T} \right) = f\left( x \right)$ for all x. If T is the smallest positive real number such that $f\left( {x + T} \right) = f\left( x \right)$ for all x, then T is called the fundamental period. We must know the periodicity of all the trigonometric functions in order to tackle such problems. One must have a strong grip over the concepts of trigonometry to solve the question.
Complete step by step answer:
So, we have to calculate the value of $\sin \left( { - \dfrac{\pi }{2}} \right)$. Now, we know that trigonometric functions are periodic functions. Since cosine and sine function are periodic functions with periods of $2\pi $, so the value of cosine and sine gets repeated after intervals in multiples of $2\pi $. Hence, we can add or subtract the term $\left( {2\pi } \right)$ in the angle of the trigonometric function according to our choice.
\[ \Rightarrow \sin \left( { - \dfrac{\pi }{2}} \right) = \sin \left( { - \dfrac{\pi }{2} + 2\pi } \right)\]
Now, simplifying the angle of the trigonometric function, we get,
\[ \Rightarrow \sin \left( { - \dfrac{\pi }{2}} \right) = \sin \left( {\dfrac{{3\pi }}{2}} \right)\]
So, splitting up the angle \[\left( {\dfrac{{3\pi }}{2}} \right)\] as \[\left( {\pi + \dfrac{\pi }{2}} \right)\], we get,
\[ \Rightarrow \sin \left( { - \dfrac{\pi }{2}} \right) = \sin \left( {\pi + \dfrac{\pi }{2}} \right)\]
Now, we also the trigonometric formula $\sin \left( {\pi + \theta } \right) = - \sin \theta $.
\[ \Rightarrow \sin \left( { - \dfrac{\pi }{2}} \right) = - \sin \left( {\dfrac{\pi }{2}} \right)\]
We know the value of \[\sin \left( {\dfrac{\pi }{2}} \right)\] is $\left( 1 \right)$. Substituting the same, we get,
$ \therefore $$ - 1$
Hence, the value of $\sin \left( { - \dfrac{\pi }{2}} \right)$ is $\left( { - 1} \right)$.
Note: Periodic Function is a function that repeats its value after a certain interval. For a real number $T > 0$, $f\left( {x + T} \right) = f\left( x \right)$ for all x. If T is the smallest positive real number such that $f\left( {x + T} \right) = f\left( x \right)$ for all x, then T is called the fundamental period. We must know the periodicity of all the trigonometric functions in order to tackle such problems. One must have a strong grip over the concepts of trigonometry to solve the question.
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