Question

What is the value of $\sin \infty$ and $\cos \infty$ ?

Answer 1

Hint: We have to find the value of $\sin \infty$ and $\cos \infty$. We will solve this question stating the intervals in which the value of sine and cosine function lies . We also give the explanation for the value of the infinity function of cosine and sine function . We can also state the periodicity of the cosine and sine function . We should also have the knowledge of ranges of trigonometric functions .

Complete step-by-step answer:
We know that , the maximum value of a \[\sin\] function is always \[1\] for every \[\left( {{\text{ }}4n{\text{ }} + {\text{ }}1{\text{ }}} \right){\text{ }} \times {\text{ }}\dfrac{\pi }{2}\] , where $n$ is a whole number
whereas the minimum value of the \[\sin\] function is always \[ - 1\] for every \[\left( {{\text{ }}4n{\text{ }} - 1{\text{ }}} \right){\text{ }} \times {\text{ }}\dfrac{\pi }{2}\] , where $n$ is a whole number .
So ,
\[ - 1{\text{ }} \leqslant {\text{ }} \sin{\text{ }}x{\text{ }} \leqslant {\text{ }}1\]
Hence , the value of \[\sin{\text{ }}x\] lies in between the interval \[\left[ {{\text{ }} - 1{\text{ }},{\text{ }}1{\text{ }}} \right]\] where $x$ is between the interval \[\left[ {{\text{ }}\dfrac{{ - \pi }}{2}{\text{ }},{\text{ }}\dfrac{\pi }{2}} \right]\]
For any of the value of $x$ it will lie in between the interval \[\left[ {{\text{ }} - 1{\text{ }},{\text{ }}1{\text{ }}} \right]\]. The exact value of $\sin \infty$ cannot be calculated but it will only lie in between the interval \[\left[ {{\text{ }} - {\text{ }}1{\text{ }},{\text{ }}1{\text{ }}} \right]{\text{ }}.\]
Similarly for $\cos \infty$
We know that , the maximum value of a \[\cos\] function is always $1$ for every \[\left( {{\text{ }}2n{\text{ }}} \right){\text{ }} \times {\text{ }}\pi \] , where $n$ is a whole number
whereas the minimum value of the \[sin\]function is always \[ - 1\] for every \[\left( {{\text{ }}2n{\text{ }} + {\text{ }}1{\text{ }}} \right){\text{ }} \times {\text{ }}\pi \] , where $n$ is a whole number .
So ,
\[ - 1{\text{ }} \leqslant {\text{ }}cos{\text{ }}x{\text{ }} \leqslant {\text{ }}1\]
Hence , the value of \[cos{\text{ }}x\] lies in between the interval \[\left[ {{\text{ }} - 1{\text{ }},{\text{ }}1{\text{ }}} \right]\] where $x$ is between the interval \[\left[ {{\text{ }}0{\text{ }},{\text{ }}\pi {\text{ }}} \right]\] .
For any of the value of $x$ it will lie in between the interval \[\left[ {{\text{ }} - 1{\text{ }},{\text{ }}1{\text{ }}} \right]\]. The exact value of $\cos \infty$ cannot be calculated but it will only lie in between the interval \[\left[ {{\text{ }} - {\text{ }}1{\text{ }},{\text{ }}1{\text{ }}} \right]\] .
Hence , the value of $\sin \infty$ and $\cos \infty$ lies in the interval \[\left[ {{\text{ }} - 1{\text{ }},{\text{ }}1{\text{ }}} \right]\].

Note: The periodic value of \[\cos\] and \[\sin\] function is \[2\pi \] I.e. the value of \[\cos\] and \[\sin\] function repeats after an interval of \[2\pi \] .
The expansions of the trigonometric terms :
\[\cos x = 1 - (\dfrac{{{x^2}}}{{2!}}) + (\dfrac{{{x^4}}}{{4!}}) - (\dfrac{{{x^6}}}{{6!}}) + .......................\]
\[\sin x = x - (\dfrac{{{x^3}}}{{3!}}) + (\dfrac{{{x^5}}}{{5!}}) - (\dfrac{{{x^7}}}{{7!}}) + .................\]