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Chapter 6: Problem 28

Evaluate \(\cos \left(\cos ^{-1} \frac{2}{3}+\tan ^{-1} 3\right)\).

Short Answer

Expert verified
The short answer is: \(\cos \left(\cos ^{-1} \frac{2}{3}+\tan ^{-1} 3\right) = \frac{20 - 9\sqrt{5}}{30}\).

Step by step solution

01

Evaluate inverse trigonometric functions

We are given \(\cos^{-1} \frac{2}{3}\) and \(\tan^{-1} 3\). We can assign their values to two variables, say A and B respectively: \(A = \cos^{-1} \frac{2}{3}\), and \(B = \tan^{-1} 3\).
02

Express A and B in terms of sine and cosine functions

Using the definitions of inverse trigonometric functions, we can express A and B in terms of sine and cosine functions: \(\cos A = \frac{2}{3}\), and \(\tan B = 3 = \frac{\sin B}{\cos B}\).
03

Find sine and cosine values for B

From the expression of \(\tan B\), we have \(\sin B = 3\cos B\). Now we can use the Pythagorean identity \(\sin^{2}B + \cos^{2}B = 1\) to find the sine and cosine values for B: \((3\cos B)^{2} + (\cos B)^{2} = 1\). \(9\cos^{2}B + \cos^{2}B = 1\). \(10\cos^{2}B = 1\). \(\cos^{2}B = \frac{1}{10}\). \(\cos B = \frac{1}{\sqrt{10}}\). We can now find \(\sin B\) using the expression we found earlier: \(\sin B = 3\cos B = \frac{3}{\sqrt{10}}\).
04

Evaluate the sum of A and B

Now, we need to find \(\cos \left(A+B\right)\). Using the cosine addition formula, we have: \(\cos \left(A+B\right) = \cos A \cos B - \sin A \sin B\).
05

Plug in values and compute the result

We plug in the values we found for \(\sin A\), \(\sin B\), \(\cos A\), and \(\cos B\): \(\cos \left(A+B\right) = \left(\frac{2}{3}\right)\left(\frac{1}{\sqrt{10}}\right) - \left(\frac{\sqrt{5}}{3}\right)\left(\frac{3}{\sqrt{10}}\right)\). After simplifying the expression, we get: \(\cos \left(A+B\right) = \frac{2}{3\sqrt{10}} - \frac{3\sqrt{5}}{10} = \frac{20 - 9\sqrt{5}}{30}\). Hence, the final answer is: \(\cos \left(\cos ^{-1} \frac{2}{3}+\tan ^{-1} 3\right) = \frac{20 - 9\sqrt{5}}{30}\).

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