- The expression \(\sin(\frac{2\pi}{3} + \frac{5\pi}{6})\) evaluates to \(\boxed{-1}\) using the sine addition identity.
- The expression \(\cos(195^{\circ})\) is calculated as \(\cos(240^{\circ}-45^{\circ})\), which equals \(\boxed{-\frac{\sqrt{2}+\sqrt{6}}{4}}\).
- The expression \(\tan(\frac{\pi}{4} + \frac{\pi}{3})\) simplifies to \(\boxed{-2-\sqrt{3}}\) via the tangent addition identity.
Explanation
- Evaluating Trigonometric Expressions
To evaluate the given expressions, we will use the standard trigonometric sum and difference formulas for sine, cosine, and tangent.
- The first expression is \(\sin\left(\frac{2\pi}{3} + \frac{5\pi}{6}\right)\). We use the sine addition formula: \(\sin(A+B) = \sin A \cos B + \cos A \sin B\). Here, \(A = \frac{2\pi}{3}\) and \(B = \frac{5\pi}{6}\).
\(\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\), \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\)
\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\), \(\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\)
Substituting these: \(\left(\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right) + \left(-\frac{1}{2}\right)\left(\frac{1}{2}\right) = -\frac{3}{4} - \frac{1}{4} = -1\).
- The second expression is \(\cos(195^{\circ})\). We write \(195^{\circ}\) as \(240^{\circ} - 45^{\circ}\) and use the cosine difference formula: \(\cos(A-B) = \cos A \cos B + \sin A \sin B\).
\(\cos(240^{\circ}) = -\frac{1}{2}\), \(\sin(240^{\circ}) = -\frac{\sqrt{3}}{2}\)
\(\cos(45^{\circ}) = \frac{\sqrt{2}}{2}\), \(\sin(45^{\circ}) = \frac{\sqrt{2}}{2}\)
Substituting these: \(\left(-\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right) + \left(-\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) = -\frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = -\frac{\sqrt{2}+\sqrt{6}}{4}\).
- The third expression is \(\tan\left(\frac{\pi}{4} + \frac{\pi}{3}\right)\). We use the tangent addition formula: \(\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\). Here, \(A = \frac{\pi}{4}\) and \(B = \frac{\pi}{3}\).
\(\tan\left(\frac{\pi}{4}\right) = 1\), \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\)
Substituting these: \(\frac{1 + \sqrt{3}}{1 - (1)(\sqrt{3})} = \frac{1 + \sqrt{3}}{1 - \sqrt{3}}\).
To simplify, multiply the numerator and denominator by the conjugate \(1 + \sqrt{3}\):
\(\frac{(1 + \sqrt{3})^2}{1 - 3} = \frac{1 + 2\sqrt{3} + 3}{-2} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}\).
- Summary of Results
Based on the derivations, the results are:
\(\sin\left(\frac{2\pi}{3} + \frac{5\pi}{6}\right) = -1\)
\(\cos(195^{\circ}) = -\frac{\sqrt{2} + \sqrt{6}}{4}\)
\(\tan\left(\frac{\pi}{4} + \frac{\pi}{3}\right) = -2 - \sqrt{3}\)
These calculations demonstrate how sum and difference formulas simplify complex trigonometric values into exact radicals.
Examples
Trigonometric sum and difference formulas are essential in fields like physics and engineering, where they help analyze wave interference patterns and signal processing. For instance, if you are determining the combined amplitude of two sound waves of different frequencies, these formulas allow you to combine the individual trigonometric functions into a single manageable expression, making it easier to calculate peak intensities. This ensures precision in designing audio equipment and telecommunications systems.