Question

Prove that \(\arcsin x=\arctan \left(\frac{x}{\sqrt{1-x^{2}}}\right),|x|<1\)

Short answer

The identity \(\arcsin x= \arctan \left(\frac{x}{\sqrt{1-x^{2}}}\right),|x|<1\) is proven by applying trigonometric identities and the properties of the inverse trigonometric functions.

Step-by-step solution

Define Given Variables

Let \(y=\arcsin x\), then by definition \(x=\sin y\). Likewise, let \(w=\arctan \left(\frac{x}{\sqrt{1-x^{2}}}\right)\), then \(\frac{x}{\sqrt{1-x^{2}}}=\tan w\).

Apply the Trigonometric Identity

We know that the trigonometric identity \(1+\tan^{2}w = \sec^{2}w\) holds. Then we substitute \(tan w = \frac{x}{\sqrt{1-x^{2}}}\) into this identity, so we get \(1 + \left(\frac{x}{\sqrt{1-x^{2}}}\right)^{2} = \sec^{2}w\). Simplifying the left-hand side gives \(\sec^{2}w = \frac{1}{1-x^{2}}\).

Use the Identity \(\sec^{2}w -1= \tan^{2}w\)

Let's substitute the value of \(\sec^{2}w\) found in step 2 into the identity \(\sec^{2}w -1= \tan^{2}w\), which simplifies to \( \frac{1}{1-x^{2}} -1= \tan^{2}w = \frac{x^2}{1-x^{2}}\). Then, solving for \(\tan^{2}w\) yields \(\tan^{2}w= \left(\frac{x}{\sqrt{1-x^{2}}}\right)^{2}\). Taking the square root of both sides gives \(\tan w =\frac{x}{\sqrt{1-x^{2}}}\), and hence \(w = \arctan \left(\frac{x}{\sqrt{1-x^{2}}}\right)\).

Conclusion

At the beginning we stated \(y=\arcsin x\), and in step 3 we showed that through substitution and simplification we arrived at \(w = \arctan \left(\frac{x}{\sqrt{1-x^{2}}}\right)\) with the same \(x = \sin y\). This shows that \(y = w\), therefore the identity \(\arcsin x=\arctan \left(\frac{x}{\sqrt{1-x^{2}}}\right),|x|<1\) is proven.