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

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

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

Answer

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}). [ \begin{align*} \sin(15^{\circ})&=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}\ &=\frac{\sqrt{6}-\sqrt{2}}{4} \end{align*} ]

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)