Question

If \(\tan A<0\) and \(\tan 2A = -\frac{4}{3}\), then \(\cos 6A\) is:

• A. \(\frac{117}{125}\)
• B. \(-\frac{117}{125}\)
• C. \(\frac{120}{169}\)
• D. \(-\frac{120}{169}\)

Hint:

In trigonometric problems involving multiple angles, work systematically through known identities, starting from given values to required angles.

Answer 1

The Correct Option is B

We found that \(\tan 2A = -\frac{4}{3}\) and \(\tan A = -\frac{1}{2}\). Since \(\tan A<0\), \(A\) is in the second or fourth quadrant. Since \(\tan 2A<0\), \(2A\) is in the second or fourth quadrant. We know that \(\cos 2A = \frac{1-\tan^2 A}{1+\tan^2 A}\). \[\cos 2A = \frac{1 - \left(-\frac{1}{2}\right)^2}{1 + \left(-\frac{1}{2}\right)^2} = \frac{1 - \frac{1}{4}}{1 + \frac{1}{4}} = \frac{\frac{3}{4}}{\frac{5}{4}} = \frac{3}{5}\] However, since \(\tan 2A = -\frac{4}{3}\) and \(\cos 2A = \frac{3}{5}\), we have \(\sin 2A = \tan 2A \cdot \cos 2A = -\frac{4}{3} \cdot \frac{3}{5} = -\frac{4}{5}\). Since \(\sin 2A<0\) and \(\cos 2A>0\), \(2A\) is in the fourth quadrant. Therefore, \(\cos 2A = \frac{3}{5}\). We want to find \(\cos 6A = \cos(3(2A))\). Let \(2A = \theta\). Then \(\cos \theta = \frac{3}{5}\). We want to find \(\cos 3\theta = 4\cos^3 \theta - 3\cos \theta\). \[\cos 6A = 4\left(\frac{3}{5}\right)^3 - 3\left(\frac{3}{5}\right)\] \[= 4\left(\frac{27}{125}\right) - \frac{9}{5}\] \[= \frac{108}{125} - \frac{225}{125}\] \[= -\frac{117}{125}\] Thus, the correct answer is \(-\frac{117}{125}\). Final Answer: The final answer is $\boxed{(2)}$