Question

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

Short answer

\(\arccos(x)=\frac{\pi}{2}-\arctan\left(\frac{x}{\sqrt{1-x^{2}}}\right),|x|<1\)

Step-by-step solution

Express Given Formulas using Pythagorean Identity

To start, one can recall the Pythagorean identity: \(1 = \cos^{2}\theta + \sin^{2}\theta\). From this identity, \( \sin(\arccos(x)) \) can be expressed as \(\sqrt{1 - x^{2}}\), since the arcsine and arccosine are complementary angles, i.e. \( \arccos(x) = \frac{\pi}{2} - \arcsin(x) \). Thus we can represent \( \sin(\arccos(x)) \) as \( \frac{x}{\sqrt{1 - x^{2}}} \).

Employing the Arctangent Function

The arctangent function, \(\arctan(x)\), is defined as the inverse of the tangent function. The tangent of an angle is the ratio of the sine of the angle to its cosine. Thus, \( \arctan\left(\frac{x}{\sqrt{1 - x^{2}}}\right) \) can be interpreted as an angle whose tangent is \( \frac{x}{\sqrt{1 - x^{2}}} \). However, we know that the tangent of an angle is also the cotangent of its complementary angle. Hence, the representation becomes \(\frac{\pi}{2} - \arccos(x)\).

Final Equality

Based on the previous steps, it's evident that \( \arccos(x) \) equals \(\frac{\pi}{2} - \arctan\left(\frac{x}{\sqrt{1 - x^{2}}}\right)\). Therefore, the initial expression has been successfully proved.