Given that \(\sin x = \frac{5}{13}\) and \(\sin y = \frac{8}{17}\), where x and y are acute, find \(\cos(x+y)\).
Answer Details
To solve the problem, we need to use the sum formula for cosine:
\[\cos(x+y) = \cos x \cos y - \sin x \sin y\]
We are given the values of \(\sin x\) and \(\sin y\), so we need to find \(\cos x\) and \(\cos y\). To do this, we use the fact that \(\sin^2 x + \cos^2 x = 1\) and \(\sin^2 y + \cos^2 y = 1\) for any angle x and y.
From \(\sin x = \frac{5}{13}\), we can find \(\cos x\) by using the Pythagorean identity:
\[\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13}\]
Similarly, from \(\sin y = \frac{8}{17}\), we can find \(\cos y\):
\[\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - \left(\frac{8}{17}\right)^2} = \frac{15}{17}\]
Now we can substitute into the formula for \(\cos(x+y)\):
\[\cos(x+y) = \cos x \cos y - \sin x \sin y = \frac{12}{13} \cdot \frac{15}{17} - \frac{5}{13} \cdot \frac{8}{17} = \frac{140}{221}\]
Therefore, the answer is \(\frac{140}{221}\).