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

Prove that sin⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} for ∣x∣≤1\left| x \right|\le 1.

Explanation

Solution

We prove this equation by using the sine to cosine transformation that is sinθ=cos⁡(π2−θ)sin\theta =\cos \left( \dfrac{\pi }{2}-\theta \right) or cos⁡θ=sin⁡(π2−θ)\cos \theta =\sin \left( \dfrac{\pi }{2}-\theta \right) where \theta $$$$\le $$$$\dfrac{\pi }{2} or there will be alternative method also to proof this basic inverse trigonometric identity which uses the eleventh class trigonometric equation which is given by: cos⁡(A−B)\cos (A-B) == \cos A$$$$\cos B\sin A\sin B

Complete step by step solution:
Let us take x=\sin \theta $$$$=\cos \left( \dfrac{\pi }{2}-\theta \right)$$$$\Rightarrow sin⁡−1x=θ{{\sin }^{-1}}x=\theta
And cos⁡−1x=π2−θ{{\cos }^{-1}}x=\dfrac{\pi }{2}-\theta
Now we will substitute the equation sin⁡−1x=θ{{\sin }^{-1}}x=\theta into cos⁡−1x=π2−θ{{\cos }^{-1}}x=\dfrac{\pi }{2}-\theta
We get, {{\cos }^{-1}}x$$$$=\dfrac{\pi }{2}$$$$-$$$${{\sin }^{-1}}x
\Rightarrow $$$${{\cos }^{-1}}x$$$$+$$$${{\sin }^{-1}}x=π2\dfrac{\pi }{2}
So, it’s proving thatsin⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}.
Here, \theta $$$$\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] this implies that x∈[−1,1]x\in \left[ -1,1 \right]
Therefore, it is valid only for∣x∣≤1\left| x \right|\le 1.
Hence Proved.

Note: Alternative Method:
This can be proved by using the formula cos⁡(A−B)=cos⁡Acos⁡Bsin⁡Asin⁡B\cos (A-B)=\cos A\cos B\sin A\sin B where A=π2A=\dfrac{\pi }{2} and B=θB=\theta
On putting A=π2A=\dfrac{\pi }{2}and B=θB=\theta
We get, cos⁡(π2−θ)\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⁡θ\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.