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Question: If \(\sin ({{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x)=1\) , then find the value of x....

If sin⁡(sin⁡−115+cos⁡−1x)=1\sin ({{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x)=1 , then find the value of x.

Explanation

Solution

To solve this inverse trigonometric question, what we will do is, we will first find the value of sin⁡−115+cos⁡−1x{{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}xin radian and then using the inverse trigonometric property which is sin⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}, we will find the value of x.

Complete step by step answer:
Now, before we start solving this question, let us see values of sin⁡x\sin xfor different degree angles and we will also see what are some inverse trigonometric formulas.
Now, value of sin⁡x\sin x at 0 is 0, value of sin⁡x\sin x at π6\dfrac{\pi }{6} is 12\dfrac{1}{2}, value of sin⁡x\sin x at π4\dfrac{\pi }{4} is 12\dfrac{1}{\sqrt{2}}, value of sin⁡x\sin x at π3\dfrac{\pi }{3} is 32\dfrac{\sqrt{3}}{2} and value of sin⁡x\sin x at π2\dfrac{\pi }{2} is 1.
Now, some inverse trigonometric identities are,
sin⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}, tan⁡−1x+cot⁡−1x=π2{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2} and sec⁡−1x+cosec⁡−1x=π2{{\sec }^{-1}}x+{{\operatorname{cosec}}^{-1}}x=\dfrac{\pi }{2}.
Now, in question it is given that sin⁡(sin⁡−1x+cos⁡−1x)=1\sin ({{\sin }^{-1}}x+{{\cos }^{-1}}x)=1
So, we can re – write sin⁡(sin⁡−1x+cos⁡−1x)=1\sin ({{\sin }^{-1}}x+{{\cos }^{-1}}x)=1 as
sin⁡(sin⁡−115+cos⁡−1x)=sin⁡(π2)\sin ({{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x)=\sin \left( \dfrac{\pi }{2} \right), as we discussed above that value of sin⁡x\sin x at π2\dfrac{\pi }{2} is 1.
As on right side and left side we have function of sin, so we can compare the inputs,
So, on comparing, we get
sin⁡−115+cos⁡−1x=π2{{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x=\dfrac{\pi }{2}
Now, we know that sin⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2},
So, comparing sin⁡−115+cos⁡−1x=π2{{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x=\dfrac{\pi }{2} with sin⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}, we get
x=15x=\dfrac{1}{5}
Hence, the value of sin⁡(sin⁡−115+cos⁡−1x)\sin ({{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x) is equal to one if x is equals to 15\dfrac{1}{5} that is sin⁡(sin⁡−115+cos⁡−1x)=1\sin ({{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x)=1for x=15x=\dfrac{1}{5}.

Note:
While solving questions based on inverse trigonometric function, we must know all the properties of inverse trigonometric function as well as trigonometric function because some questions are tricky, which can be solved easily with help of these identities. While solving the questions and evaluating values of x, try to avoid calculation error as this will make you stuck in between of the solution or may give you incorrect answers.