Question

Find the principal value of \[{{\sec }^{-1}}\left( 2\sin \dfrac{3\pi }{4} \right)\].

Answer 1

Hint:For the above question we have to know about the principal values of an inverse trigonometric function. Principal value of an inverse trigonometric function is a value that belongs to the principal branch of the range of the function. We know that the principal branch of range of \[{{\sec }^{-1}}x\] is \[\left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}\].

Complete step-by-step answer:
We have been given the expression \[{{\sec }^{-1}}\left( 2\sin \dfrac{3\pi }{4} \right)\].
Now we know that \[\sin \dfrac{3\pi }{4}=\dfrac{1}{\sqrt{2}}\].
On substituting the value of \[\sin \dfrac{3\pi }{4}\] in the given expression we get as follows:
\[{{\sec }^{-1}}\left( 2\sin \dfrac{3\pi }{4} \right)={{\sec }^{-1}}\left( 2\times \dfrac{1}{\sqrt{2}} \right)={{\sec }^{-1}}\left( 2\times \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}} \right)={{\sec }^{-1}}\left( 2\times \dfrac{\sqrt{2}}{2} \right)={{\sec }^{-1}}\left( \sqrt{2} \right)\]
Now we know that the principal value means the value which lies between the defined range of the function.
Now we have \[{{\sec }^{-1}}\left( 2\sin \dfrac{3\pi }{4} \right)={{\sec }^{-1}}\left( \sqrt{2} \right)\].
Since we know that the value of \[\sec \left( \dfrac{\pi }{4} \right)=\sqrt{2}\]
So by substituting the value of \[\sqrt{2}\] in the given expression, we get as follows:
\[{{\sec }^{-1}}\left( 2\sin \dfrac{3\pi }{4} \right)={{\sec }^{-1}}\left[ \sec \left( \dfrac{\pi }{4} \right) \right]\]
We know the relation that \[{{\sec }^{-1}}\sec \theta =\theta \] where \[\theta \] must lie between \[\left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}\].
So, we can write that the value of \[{{\sec }^{-1}}\left( 2\sin \dfrac{3\pi }{4} \right)=\dfrac{\pi }{4}\]
Therefore the principal value of \[{{\sec }^{-1}}\left( 2\sin \dfrac{3\pi }{4} \right)\] is equal to \[\dfrac{\pi }{4}\].

Note: Be careful while finding the principal value of inverse trigonometric functions and do check it once that the value must lie between the principal branch of the range of the function. Also be careful while finding the values of \[sin\dfrac{3\pi }{4}\], by mistake we may write \[sin\dfrac{3\pi }{4}=\dfrac{-1}{\sqrt{2}}\] which is wrong since \[sin\dfrac{3\pi }{4}=\dfrac{1}{\sqrt{2}}\] , because \[sin\left( \pi -\dfrac{\pi }{4} \right)=\sin \dfrac{\pi }{4}\] not \[\left( -\sin \dfrac{\pi }{4} \right)\]. So, this type of mistake must be avoided.