Problem 22 - Calculus of a Single Variable

Question

Evaluate each expression without using a calculator. (Hint: See Example 3.) (a) \(\tan \left(\arccos \frac{\sqrt{2}}{2}\right)\) (b) \(\cos \left(\arcsin \frac{5}{13}\right)\)

Step-by-Step Solution

Answer

(a) The value of the expression \(\tan \left(\arccos \frac{\sqrt{2}}{2}\right)\) is 1. \n(b) The value of the expression \(\cos \left(\arcsin \frac{5}{13}\right)\) is \(\frac{12}{13}\).

Step 1: Evaluate Expression (a)

Expression (a) is \(\tan \left(\arccos \frac{\sqrt{2}}{2}\right)\). It can be evaluated by realising that \(\arccos \frac{\sqrt{2}}{2}\) is equal to \(\frac{\pi}{4}\) because \(\cos( \frac{\pi}{4}) = \frac{\sqrt{2}}{2}\). This implies that the expression can be simplified to \(\tan(\frac{\pi}{4})\), because of the inverse cosine transformation. The tangent of \(\frac{\pi}{4}\) equals 1.

Step 2: Evaluate Expression (b)

Expression (b) is \(\cos \left(\arcsin \frac{5}{13}\right)\). It can be evaluated by using the Pythagorean trigonometric identity: \( \sin^2(x) + \cos^2(x) = 1 \). Given that \(\sin(x) = \frac{5}{13}\), we can solve for \(\cos(x)\) to obtain \(\cos(x) = \sqrt{1 - \sin^2(x)} = \sqrt{1-\left(\frac{5}{13}\right)^2} = \sqrt{1-\frac{25}{169}} = \frac{12}{13}\). So the value of the expression is \(\cos \left(\arcsin \frac{5}{13}\right) = \frac{12}{13}\). Note here that we assumed the angle to be in the first quadrant, hence positive cosine value.