Question

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

Answer 1

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

Complete step by step answer:
To Prove : $\arcsin x + \arccos x = \dfrac{\pi }{2}$ or $si{n^{ - 1}}x + co{s^{ - 1}}x = \dfrac{\pi }{2}$
Proof:
Let assume $si{n^{ - 1}}x = \theta $, ------(1)
Since we know that $x = \sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right)$
We can say that
$
\Rightarrow x = \cos \left( {\dfrac{\pi }{2} - \theta } \right) \\
   \Rightarrow {\cos ^{ - 1}}x = \left( {\dfrac{\pi }{2} - \theta } \right) \\
 $
Substituting value of $\theta $from what we have assumed in (1)
\[
   \Rightarrow {\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}x \\
   \Rightarrow {\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2} \\
 \]
Therefore LHS = RHS. Hence, Proved.

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

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