How would you find the exact value of the six trigonometric functions of $\dfrac{{5\pi }}{2}$ ?
Answer
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Hint:In this question, we have to find the value of all the trigonometric functions which are sine, cosine, tangent, cotangent, cosecant and secant for the $\dfrac{{5\pi }}{2}$. Firstly, know that the $\dfrac{{5\pi }}{2} = 2\pi + \dfrac{\pi }{2}$ . Now, $2\pi $is the angle where the trigonometric functions do not change. And, hence finding the values.
Complete step by step answer:
In this respective question, firstly we are splitting the angle given into those values whose values are learned by everybody which are in the form of $\pi ,2\pi \ldots $ and so on.So, writing $\dfrac{{5\pi }}{2}$ as $2\pi + \dfrac{\pi }{2}$. Make sure that the quadrant now exists to be first where all trigonometric functions are positive. And also, at $\pi ,2\pi \ldots $ the trigonometric functions do not change. Trigonometric functions would change at $\dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2} \ldots $
The functions change as an example: $\sin \to \cos $,$\cos ec \to \sec ,\tan \to \cot $ and vice-versa.Finding the values
$\sin \left( {2\pi + \dfrac{\pi }{2}} \right) = \sin \left( {\dfrac{\pi }{2}} \right) = 1$
$\Rightarrow\cos \left( {2\pi + \dfrac{\pi }{2}} \right) = \cos \left( {\dfrac{\pi }{2}} \right) = 0$
$
\Rightarrow\tan \left( {2\pi + \dfrac{\pi }{2}} \right) = \tan \left( {\dfrac{\pi }{2}} \right) = \infty \\
\Rightarrow\cot \left( {2\pi + \dfrac{\pi }{2}} \right) = \cot \left( {\dfrac{\pi }{2}} \right) = 0 \\
\Rightarrow\cos ec\left( {2\pi + \dfrac{\pi }{2}} \right) = \cos ec\left( {\dfrac{\pi }{2}} \right) = 1 \\
\therefore\sec \left( {2\pi + \dfrac{\pi }{2}} \right) = \sec \left( {\dfrac{\pi }{2}} \right) = \infty \\
$
These are the values for the given angle for all the trigonometric functions.
Note:Make sure you have learnt the full trigonometric table. The quadrants should be observed carefully. Learn which functions are positive in which quadrant. In the first quadrant, all functions are positive. In the second, sine and cosecant are positive, else negative. In the third quadrant, tangent and cotangent are positive. In fourth, cosecant and secant are positive.
Complete step by step answer:
In this respective question, firstly we are splitting the angle given into those values whose values are learned by everybody which are in the form of $\pi ,2\pi \ldots $ and so on.So, writing $\dfrac{{5\pi }}{2}$ as $2\pi + \dfrac{\pi }{2}$. Make sure that the quadrant now exists to be first where all trigonometric functions are positive. And also, at $\pi ,2\pi \ldots $ the trigonometric functions do not change. Trigonometric functions would change at $\dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2} \ldots $
The functions change as an example: $\sin \to \cos $,$\cos ec \to \sec ,\tan \to \cot $ and vice-versa.Finding the values
$\sin \left( {2\pi + \dfrac{\pi }{2}} \right) = \sin \left( {\dfrac{\pi }{2}} \right) = 1$
$\Rightarrow\cos \left( {2\pi + \dfrac{\pi }{2}} \right) = \cos \left( {\dfrac{\pi }{2}} \right) = 0$
$
\Rightarrow\tan \left( {2\pi + \dfrac{\pi }{2}} \right) = \tan \left( {\dfrac{\pi }{2}} \right) = \infty \\
\Rightarrow\cot \left( {2\pi + \dfrac{\pi }{2}} \right) = \cot \left( {\dfrac{\pi }{2}} \right) = 0 \\
\Rightarrow\cos ec\left( {2\pi + \dfrac{\pi }{2}} \right) = \cos ec\left( {\dfrac{\pi }{2}} \right) = 1 \\
\therefore\sec \left( {2\pi + \dfrac{\pi }{2}} \right) = \sec \left( {\dfrac{\pi }{2}} \right) = \infty \\
$
These are the values for the given angle for all the trigonometric functions.
Note:Make sure you have learnt the full trigonometric table. The quadrants should be observed carefully. Learn which functions are positive in which quadrant. In the first quadrant, all functions are positive. In the second, sine and cosecant are positive, else negative. In the third quadrant, tangent and cotangent are positive. In fourth, cosecant and secant are positive.
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