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# How would you find the exact value of the six trigonometric functions of $\dfrac{{5\pi }}{2}$ ?

Last updated date: 29th Feb 2024
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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.

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
$\sin \left( {2\pi + \dfrac{\pi }{2}} \right) = \sin \left( {\dfrac{\pi }{2}} \right) = 1$
$\Rightarrow\cos \left( {2\pi + \dfrac{\pi }{2}} \right) = \cos \left( {\dfrac{\pi }{2}} \right) = 0$
$\Rightarrow\tan \left( {2\pi + \dfrac{\pi }{2}} \right) = \tan \left( {\dfrac{\pi }{2}} \right) = \infty \\ \Rightarrow\cot \left( {2\pi + \dfrac{\pi }{2}} \right) = \cot \left( {\dfrac{\pi }{2}} \right) = 0 \\ \Rightarrow\cos ec\left( {2\pi + \dfrac{\pi }{2}} \right) = \cos ec\left( {\dfrac{\pi }{2}} \right) = 1 \\ \therefore\sec \left( {2\pi + \dfrac{\pi }{2}} \right) = \sec \left( {\dfrac{\pi }{2}} \right) = \infty \\$