Question

Prove that \[{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}\] for \[\left| x \right|\le 1\].

Answer 1

Hint: We prove this equation by using the sine to cosine transformation that is \[sin\theta =\cos \left( \dfrac{\pi }{2}-\theta \right)\] or \[\cos \theta =\sin \left( \dfrac{\pi }{2}-\theta \right)\] where \[\theta \]\[\le \]\[\dfrac{\pi }{2}\] or there will be alternative method also to proof this basic inverse trigonometric identity which uses the eleventh class trigonometric equation which is given by: \[\cos (A-B)\] \[=\] \[\cos A\]\[\cos B\sin A\sin B\]

Complete step by step solution:
Let us take \[x=\sin \theta \]\[=\cos \left( \dfrac{\pi }{2}-\theta \right)\]\[\Rightarrow \] \[{{\sin }^{-1}}x=\theta \]
And \[{{\cos }^{-1}}x=\dfrac{\pi }{2}-\theta \]
Now we will substitute the equation \[{{\sin }^{-1}}x=\theta \] into \[{{\cos }^{-1}}x=\dfrac{\pi }{2}-\theta \]
We get, \[{{\cos }^{-1}}x\]\[=\dfrac{\pi }{2}\]\[-\]\[{{\sin }^{-1}}x\]
\[\Rightarrow \]\[{{\cos }^{-1}}x\]\[+\]\[{{\sin }^{-1}}x\]=\[\dfrac{\pi }{2}\]
So, it’s proving that\[{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}\].
Here, \[\theta \]\[\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\] this implies that \[x\in \left[ -1,1 \right]\]
Therefore, it is valid only for\[\left| x \right|\le 1\].
Hence Proved.

Note: Alternative Method:
This can be proved by using the formula \[\cos (A-B)=\cos A\cos B\sin A\sin B\] where \[A=\dfrac{\pi }{2}\] and \[B=\theta \]
On putting \[A=\dfrac{\pi }{2}\]and \[B=\theta \]
We get, \[\cos (\dfrac{\pi }{2}-\theta )\]=\[\cos \dfrac{\pi }{2}\]\[\cos \theta \]\[+\]\[\sin \dfrac{\pi }{2}\]\[\sin \theta \]= \[0\]\[\times \]\[\cos \theta \]\[+\]\[1\]\[\times \]\[\sin \theta \]
\[\Rightarrow \]\[\sin \theta \]
Now, again follow the same steps by taking the \[\sin \theta \] is equal to x and then use the trigonometric transform identities and then substitute the value of x into any transform identities then this inverse trigonometric basic identity will be obtained.