To find the value of \(\cos(105^\circ)\), we break down \(105^\circ\) into the sum of two standard angles: \(105^\circ = 60^\circ + 45^\circ\). This allows us to use trigonometric identities.
The angle addition identity for cosine states: \(\cos(A + B) = \cos A \cos B - \sin A \sin B\)
Applying this identity with \(A = 60^\circ\) and \(B = 45^\circ\): \(\cos(105^\circ) = \cos(60^\circ + 45^\circ) = \cos(60^\circ)\cos(45^\circ) - \sin(60^\circ)\sin(45^\circ)\)
We use the known values for these standard angles:
Substitute these values into the expanded formula:
\(\cos(105^\circ) = \left(\frac{1}{2}\right) \left(\frac{\sqrt{2}}{2}\right) - \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{2}}{2}\right)\)
Simplify the terms:
\(\cos(105^\circ) = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}\)
Combine the terms over the common denominator:
\(\cos(105^\circ) = \frac{\sqrt{2} - \sqrt{6}}{4}\)
The calculated value is \(\frac{\sqrt{2} - \sqrt{6}}{4}\). Let's verify Option A: \(\frac{(1-\sqrt{3})}{2\sqrt{2}}\). To compare, we rationalize the denominator of Option A:
\(\frac{(1-\sqrt{3})}{2\sqrt{2}} = \frac{(1-\sqrt{3}) \times \sqrt{2}}{(2\sqrt{2}) \times \sqrt{2}} = \frac{\sqrt{2} - \sqrt{6}}{2 \times 2} = \frac{\sqrt{2} - \sqrt{6}}{4}\)
The calculated value matches Option A.
What is \(\tan \alpha + \tan \beta\) equal to?