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Question: Fill in the blank: \({{\sin }^{-1}}x+{{\cos }^{-1}}x=\\_\\_\\_\\_\\_\\_\\_\) ....

Fill in the blank: {{\sin }^{-1}}x+{{\cos }^{-1}}x=\\_\\_\\_\\_\\_\\_\\_ .

Explanation

Solution

To find the value of sin⁡−1x+cos⁡−1x{{\sin }^{-1}}x+{{\cos }^{-1}}x , we will first create an equation by equating sin⁡−1x{{\sin }^{-1}}x to θ\theta . Then from this equation, we will find x which will be equal to sin⁡θ\sin \theta . We will then use the formula cos⁡(π2−θ)=sin⁡θ\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta and substitute this in the previous equation. After a few rearrangements and substitutions, we will get the required solution.

Complete step-by-step solution:
We have to find the value of sin⁡−1x+cos⁡−1x{{\sin }^{-1}}x+{{\cos }^{-1}}x . Let us consider
sin⁡−1x=θ...(i){{\sin }^{-1}}x=\theta ...\left( i \right)
From the above equation, we can write,
x=sin⁡θx=\sin \theta
We know that cos⁡(π2−θ)=sin⁡θ\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta . Therefore, we can write the above equation as
⇒x=cos⁡(π2−θ)\Rightarrow x=\cos \left( \dfrac{\pi }{2}-\theta \right)
We can take the cos to the LHS, so that the above equation can be written as
⇒cos⁡−1x=(π2−θ)\Rightarrow {{\cos }^{-1}}x=\left( \dfrac{\pi }{2}-\theta \right)
Let us substitute equation (i) in the above equation. Hence, we will get
⇒cos⁡−1x=(π2−sin⁡−1x)\Rightarrow {{\cos }^{-1}}x=\left( \dfrac{\pi }{2}-{{\sin }^{-1}}x \right)
Now, we have to move sin⁡−1x{{\sin }^{-1}}x to the LHS. Then, we can write the above equation as
⇒cos⁡−1x+sin⁡−1x=π2\Rightarrow {{\cos }^{-1}}x+{{\sin }^{-1}}x=\dfrac{\pi }{2}
Therefore, sin⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} .

Note: Students must be thorough with all the rules , formulas and properties of trigonometric functions. They must also be thorough with the arc functions or inverse trigonometric functions. The inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. We usually call inverse trigonometric functions as arc functions, for example, for sin⁡−1x{{\sin }^{-1}}x we can write arcsin, for cos⁡−1x{{\cos }^{-1}}x , we can write arcos and so on. Students must know to solve algebraic equations and associated rules and properties.
The above proved property is applicable for x in the range −1≤x≤1-1\le x\le 1 or for x∈[−1,1]x\in \left[ -1,1 \right] .