Question

Given tan(A) = (3/4), 0 < A < (π/2) and cos(B) = (5/13), (3π/2) < B < 2π) determine cos(2A)

Answer 1

Given $\tan(A) = (3/4)$, $0 < A < (\pi/2)$ and $\cos(B) = (5/13)$, $(3\pi/2) < B < 2\pi)$ determine $\cos(2A)$

Solution

Let find $\cos A$. We will use formula:

$$1 + \tan^2 \alpha = \frac{1}{\cos^2 \alpha}$$ $$\cos^2 \alpha = \frac{1}{1 + \tan^2 \alpha}$$ $$\cos^2 A = \frac{1}{1 + \left(\frac{3}{4}\right)^2} = \frac{1}{1 + \frac{9}{16}} = \frac{1}{\frac{16 + 9}{16}} = \frac{16}{25}$$ $$\cos A = \pm \frac{4}{5}$$

If $0 < A < \frac{\pi}{2}$ then $\cos A = \frac{4}{5}$

Let find $\sin A$

We will use the formula:

$$\sin^2 \alpha + \cos^2 \alpha = 1$$ $$\sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{4}{5}\right)^2 = \frac{25 - 16}{25} = \frac{9}{25}$$ $$\sin A = \pm \frac{3}{5}$$

If $0 < A < \frac{\pi}{2}$ then $\sin A = \frac{3}{5}$

Let find $\cos 2A$

We will use the formula:

$$\cos 2 \alpha = \cos^2 \alpha - \sin^2 \alpha$$ $$\cos 2A = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$$

Answer $\cos 2A = \frac{7}{25}$

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