use a sum or difference formula to find the exact value of the trigonometric function. \n\n\\( \\sin \\frac…

use a sum or difference formula to find the exact value of the trigonometric function. \n\n\\( \\sin \\frac { \\pi } { 12 } \\)\n\n\\( \\sin \\frac { \\pi } { 12 } = \\square \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

use a sum or difference formula to find the exact value of the trigonometric function. \n\n\\( \\sin \\frac { \\pi } { 12 } \\)\n\n\\( \\sin \\frac { \\pi } { 12 } = \\square \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Answer

Explanation:

Step1: Express (\frac{\pi}{12}) as a difference

We know that (\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}).

Step2: Use the sine - difference formula

The sine - difference formula is (\sin(A - B)=\sin A\cos B-\cos A\sin B). Here (A = \frac{\pi}{3}) and (B=\frac{\pi}{4}). We know that (\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}), (\cos\frac{\pi}{3}=\frac{1}{2}), (\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}), (\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}). Substitute these values into the formula: [ \begin{align*} \sin\left(\frac{\pi}{3}-\frac{\pi}{4}\right)&=\sin\frac{\pi}{3}\cos\frac{\pi}{4}-\cos\frac{\pi}{3}\sin\frac{\pi}{4}\ &=\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}-\frac{1}{2}\times\frac{\sqrt{2}}{2}\ &=\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}\ &=\frac{\sqrt{6}-\sqrt{2}}{4} \end{align*} ]

Answer:

(\frac{\sqrt{6}-\sqrt{2}}{4})