- Apply the cosine subtraction formula \(\cos(45^{\circ} - 30^{\circ})\) to express \(\cos 15^{\circ}\) in terms of known trigonometric values.
- Substitute the exact values \(\cos 45^{\circ} = \frac{\sqrt{2}}{2}\), \(\cos 30^{\circ} = \frac{\sqrt{3}}{2}\), \(\sin 45^{\circ} = \frac{\sqrt{2}}{2}\), and \(\sin 30^{\circ} = \frac{1}{2}\).
- Simplify the expression to \(\frac{\sqrt{6} + \sqrt{2}}{4}\).
- Factor the expression to \(\frac{\sqrt{2}(\sqrt{3} + 1)}{4}\) to obtain \(A=2, B=3, C=1\), so the final results are \(A=\boxed{2}, B=\boxed{3}, C=\boxed{1}\).
Explanation
- Setting up the Formula
To find the exact value of \(\cos 15^{\circ}\), we use the cosine subtraction formula. We know that \(15^{\circ} = 45^{\circ} - 30^{\circ}\). The cosine subtraction formula is given by:
\[\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\]
Setting \(\alpha = 45^{\circ}\) and \(\beta = 30^{\circ}\), we have:
\[\cos 15^{\circ} = \cos(45^{\circ} - 30^{\circ}) = \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ}\]
- Calculating the Exact Value
We use the known exact values for the trigonometric functions:
\(\cos 45^{\circ} = \frac{\sqrt{2}}{2}\)
\(\cos 30^{\circ} = \frac{\sqrt{3}}{2}\)
\(\sin 45^{\circ} = \frac{\sqrt{2}}{2}\)
\(\sin 30^{\circ} = \frac{1}{2}\)
Substituting these into our formula, we get:
\[\cos 15^{\circ} = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}\]
- Identifying Constants A, B, and C
We need to express the result in the form \(\frac{\sqrt{A}(\sqrt{B} + C)}{4}\).
Comparing \(\frac{\sqrt{6} + \sqrt{2}}{4}\) with this form, we can factor \(\sqrt{2}\) out of the numerator:
\[\frac{\sqrt{2}(\sqrt{3} + 1)}{4}\]
Matching this to \(\frac{\sqrt{A}(\sqrt{B} + C)}{4}\), we identify:
\(A = 2\)
\(B = 3\)
\(C = 1\)
Examples
In engineering and physics, calculating exact values of trigonometric functions is crucial when dealing with wave propagation or mechanical rotations. For instance, if you are designing a gear system or an antenna array that rotates by 15-degree increments, having precise mathematical values ensures that the mechanical components align perfectly or that signals are transmitted in the correct phase. Using exact values instead of decimal approximations prevents cumulative errors in high-precision manufacturing or complex modeling simulations.