Question

what is the exact value of \\( \sin ( 345 ^ { \circ } ) \\)?
\\( \frac { \sqrt { 3 } + \sqrt { 2 } } { 2 } \\)
\\( \frac { \sqrt { 6 } + \sqrt { 2 } } { 4 } \\)
\\( \frac { - \sqrt { 3 } + \sqrt { 2 } } { 2 } \\)
\\( \frac { - \sqrt { 6 } + \sqrt { 2 } } { 4 } \\)

Explanation:

Step1: Express \(345^{\circ}\) as a difference of angles

\(345^{\circ}=360^{\circ} - 15^{\circ}\), so \(\sin(345^{\circ})=\sin(360^{\circ}-15^{\circ})\). Using the formula \(\sin(A - B)=\sin A\cos B-\cos A\sin B\) with \(A = 360^{\circ}\), \(B=15^{\circ}\), we know \(\sin(360^{\circ}) = 0\), \(\cos(360^{\circ})=1\), so \(\sin(345^{\circ})=-\sin(15^{\circ})\). Also, \(15^{\circ}=45^{\circ}-30^{\circ}\).

Step2: Use the sine - difference formula

By the formula \(\sin(A - B)=\sin A\cos B-\cos A\sin B\) with \(A = 45^{\circ}\), \(B = 30^{\circ}\), \(\sin(15^{\circ})=\sin(45^{\circ}-30^{\circ})=\sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}\).
Since \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\), \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\), \(\sin30^{\circ}=\frac{1}{2}\).

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Step3: Find \(\sin(345^{\circ})\)

Since \(\sin(345^{\circ})=-\sin(15^{\circ})\), then \(\sin(345^{\circ})=-\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{-\sqrt{6}+\sqrt{2}}{4}\)

Answer:

\(\frac{-\sqrt{6}+\sqrt{2}}{4}\) (corresponding to the fourth option)