write the expression as a function of x, with no angle measure involved.\n\n\\( \\sin \\left( \\frac { \\pi…

write the expression as a function of x, with no angle measure involved.\n\n\\( \\sin \\left( \\frac { \\pi } { 6 } - x \\right) \\)\n\n\\( \\sin \\left( \\frac { \\pi } { 6 } - x \\right) = \\square \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

write the expression as a function of x, with no angle measure involved.\n\n\\( \\sin \\left( \\frac { \\pi } { 6 } - x \\right) \\)\n\n\\( \\sin \\left( \\frac { \\pi } { 6 } - x \\right) = \\square \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Answer

Explanation:

Step1: Use the sine subtraction formula

The formula for (\sin(A - B)=\sin A\cos B-\cos A\sin B). Here (A = \frac{\pi}{6}) and (B=x). So, (\sin\left(\frac{\pi}{6}-x\right)=\sin\frac{\pi}{6}\cos x-\cos\frac{\pi}{6}\sin x).

Step2: Substitute the values of (\sin\frac{\pi}{6}) and (\cos\frac{\pi}{6})

We know that (\sin\frac{\pi}{6}=\frac{1}{2}) and (\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}). Substituting these values, we get (\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x).

Answer:

(\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x)