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

Find the principal value of sec⁡−1(2sin⁡3π4){{\sec }^{-1}}\left( 2\sin \dfrac{3\pi }{4} \right).

Explanation

Solution

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⁡−1x{{\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(2sin⁡3π4){{\sec }^{-1}}\left( 2\sin \dfrac{3\pi }{4} \right).
Now we know that sin⁡3π4=12\sin \dfrac{3\pi }{4}=\dfrac{1}{\sqrt{2}}.
On substituting the value of sin⁡3π4\sin \dfrac{3\pi }{4} in the given expression we get as follows:
sec⁡−1(2sin⁡3π4)=sec⁡−1(2×12)=sec⁡−1(2×12×22)=sec⁡−1(2×22)=sec⁡−1(2){{\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(2sin⁡3π4)=sec⁡−1(2){{\sec }^{-1}}\left( 2\sin \dfrac{3\pi }{4} \right)={{\sec }^{-1}}\left( \sqrt{2} \right).
Since we know that the value of sec⁡(π4)=2\sec \left( \dfrac{\pi }{4} \right)=\sqrt{2}
So by substituting the value of 2\sqrt{2} in the given expression, we get as follows:
sec⁡−1(2sin⁡3π4)=sec⁡−1[sec⁡(π4)]{{\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⁡−1sec⁡θ=θ{{\sec }^{-1}}\sec \theta =\theta where θ\theta must lie between \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\}.
So, we can write that the value of sec⁡−1(2sin⁡3π4)=π4{{\sec }^{-1}}\left( 2\sin \dfrac{3\pi }{4} \right)=\dfrac{\pi }{4}
Therefore the principal value of sec⁡−1(2sin⁡3π4){{\sec }^{-1}}\left( 2\sin \dfrac{3\pi }{4} \right) is equal to π4\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 sin3π4sin\dfrac{3\pi }{4}, by mistake we may write sin3π4=−12sin\dfrac{3\pi }{4}=\dfrac{-1}{\sqrt{2}} which is wrong since sin3π4=12sin\dfrac{3\pi }{4}=\dfrac{1}{\sqrt{2}} , because sin(π−π4)=sin⁡π4sin\left( \pi -\dfrac{\pi }{4} \right)=\sin \dfrac{\pi }{4} not (−sin⁡π4)\left( -\sin \dfrac{\pi }{4} \right). So, this type of mistake must be avoided.