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Question: How do you prove \(\arcsin x + \arccos x = \dfrac{\pi }{2}\)....

How do you prove arcsin⁡x+arccos⁡x=π2\arcsin x + \arccos x = \dfrac{\pi }{2}.

Explanation

Solution

In order to prove the above statement, we have to first assume that sin−1x=θsi{n^{ - 1}}x = \theta , so by this we can write x=sin⁡θx = \sin \theta and now using the rule of trigonometry that sin⁡θ=cos⁡(π2−θ)\sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right) which also says x=cos⁡(π2−θ)x = \cos \left( {\dfrac{\pi }{2} - \theta } \right). By using this result, take the inverse of cosine on both sides of the equation and replace the values of θ=sin⁡−1x\theta = {\sin ^{ - 1}}x to get your desired result.

Complete step by step answer:
To Prove : arcsin⁡x+arccos⁡x=π2\arcsin x + \arccos x = \dfrac{\pi }{2} or sin−1x+cos−1x=π2si{n^{ - 1}}x + co{s^{ - 1}}x = \dfrac{\pi }{2}
Proof:
Let assume sin−1x=θsi{n^{ - 1}}x = \theta , ------(1)
Since we know that x=sin⁡θ=cos⁡(π2−θ)x = \sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right)
We can say that
⇒x=cos⁡(π2−θ) ⇒cos⁡−1x=(π2−θ)  \Rightarrow x = \cos \left( {\dfrac{\pi }{2} - \theta } \right) \\\ \Rightarrow {\cos ^{ - 1}}x = \left( {\dfrac{\pi }{2} - \theta } \right) \\\
Substituting value of θ\theta from what we have assumed in (1)

⇒cos⁡−1x=π2−sin⁡−1x ⇒sin⁡−1x+cos⁡−1x=π2  \Rightarrow {\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}x \\\ \Rightarrow {\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2} \\\

Therefore LHS = RHS. Hence, Proved.

Additional information:
1. Periodic Function = A function f(x)f(x) is said to be a periodic function if there exists a real number T > 0 such that f(x+T)=f(x)f(x + T) = f(x) for all x.
If T is the smallest positive real number such that f(x+T)=f(x)f(x + T) = f(x) for all x, then T is called the fundamental period of f(x)f(x) .
Since sin⁡ (2nπ+θ)=sin⁡θ\sin \,(2n\pi + \theta ) = \sin \theta for all values of θ\theta and n∈\inN.
2. Even Function – A function f(x)f(x) is said to be an even function, if f(−x)=f(x)f( - x) = f(x) for all x in its domain.
Odd Function – A function f(x)f(x) is said to be an even function, if f(−x)=−f(x)f( - x) = - f(x) for all x in its domain.
We know that sin⁡(−θ)=−sin⁡θ.cos⁡(−θ)=cos⁡θ and tan⁡(−θ)=−tan⁡θ\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta
Therefore, sin⁡θ\sin \theta and tan⁡θ\tan \theta and their reciprocals, cos⁡ecθ\cos ec\theta and cot⁡θ\cot \theta are odd functions whereas cos⁡θ\cos \theta and its reciprocal sec⁡θ\sec \theta are even functions.

Note: Range and domain of function arcsin⁡x\arcsin x is [−π2,π2]\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] and [−1,1]\left[ { - 1,1} \right] respectively.
Range and domain of the function arccos⁡x\arccos x is [0,π]\left[ {0,\pi } \right] and [−1,1]\left[ { - 1,1} \right] respectively.