Question

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

Answer 1

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

Solution

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

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

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

Let find $\sin A$

We will use the formula:

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

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

Let find $\sin B$

We will use the formula:

$$\sin^2 B + \cos^2 B = 1$$ $$\sin^2 B = 1 - \cos^2 B = 1 - \left(\frac{5}{13}\right)^2 = \frac{169 - 25}{169} = \frac{144}{169}$$ $$\sin B = \pm \frac{12}{13}$$

If $\frac{3\pi}{2} < B < 2\pi$ then $\sin B = -\frac{12}{13}$

Let find $\sin(A-B)$. We will use the formula:

$\sin(A-B) = \sin A \cos B - \cos A \sin B$

$$\sin(A-B) = \frac{3}{5} \cdot \frac{5}{13} - \frac{4}{5} \cdot \left(-\frac{12}{13}\right) = \frac{3}{13} + \frac{48}{65} = \frac{15 + 48}{65} = \frac{63}{65}$$

Answer

$$\sin(A-B) = \frac{63}{65}$$

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