in the following problem, the expression is the right side of the formula for \\( \\cos ( \\alpha - \\beta )…

in the following problem, the expression is the right side of the formula for \\( \\cos ( \\alpha - \\beta ) \\) with particular values for \\( \\alpha \\) and \\( \\beta \\).\n\\ \\cos \\left( \\frac { 11 \\pi } { 12 } \\right) \\cos \\left( \\frac { 2 \\pi } { 3 } \\right) + \\sin \\left( \\frac { 11 \\pi } { 12 } \\right) \\sin \\left( \\frac { 2 \\pi } { 3 } \\right) \\\n\n a. identify \\( \\alpha \\) and \\( \\beta \\) in the expression.\nthe value of \\( \\alpha \\) is \\( \\frac { 11 \\pi } { 12 } \\).\nthe value of \\( \\beta \\) is \\( \\frac { 2 \\pi } { 3 } \\).\n b. write the expression as the cosine of an angle.\nthe expression is \\( \\cos \\square \\).

in the following problem, the expression is the right side of the formula for \\( \\cos ( \\alpha - \\beta ) \\) with particular values for \\( \\alpha \\) and \\( \\beta \\).\n\\ \\cos \\left( \\frac { 11 \\pi } { 12 } \\right) \\cos \\left( \\frac { 2 \\pi } { 3 } \\right) + \\sin \\left( \\frac { 11 \\pi } { 12 } \\right) \\sin \\left( \\frac { 2 \\pi } { 3 } \\right) \\\n\n a. identify \\( \\alpha \\) and \\( \\beta \\) in the expression.\nthe value of \\( \\alpha \\) is \\( \\frac { 11 \\pi } { 12 } \\).\nthe value of \\( \\beta \\) is \\( \\frac { 2 \\pi } { 3 } \\).\n b. write the expression as the cosine of an angle.\nthe expression is \\( \\cos \\square \\).

Answer

Explanation:

Step1: Use the formula $\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$

We are given the expression $\cos\left(\frac{11\pi}{12}\right)\cos\left(\frac{2\pi}{3}\right)+\sin\left(\frac{11\pi}{12}\right)\sin\left(\frac{2\pi}{3}\right)$. Comparing it with the formula $\cos(\alpha - \beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$, where $\alpha=\frac{11\pi}{12}$ and $\beta = \frac{2\pi}{3}$.

Step2: Calculate $\alpha-\beta$

Substitute $\alpha=\frac{11\pi}{12}$ and $\beta=\frac{2\pi}{3}$ into $\alpha-\beta$. First, rewrite $\frac{2\pi}{3}$ with a denominator of 12: $\frac{2\pi}{3}=\frac{2\pi\times4}{3\times4}=\frac{8\pi}{12}$. Then $\alpha-\beta=\frac{11\pi}{12}-\frac{8\pi}{12}$. Using the formula for subtracting fractions $\frac{a}{c}-\frac{b}{c}=\frac{a - b}{c}$, we have $\frac{11\pi-8\pi}{12}=\frac{3\pi}{12}=\frac{\pi}{4}$.

Answer:

The expression is $\cos\left(\frac{\pi}{4}\right)$