find the exact value of the following expression. sin 11π/12 rewrite the expression using a sum or…

find the exact value of the following expression. sin 11π/12 rewrite the expression using a sum or difference formula: choose the correct one. a. sin 11π/12 = sin(2π/3 - π/4) = cos 2π/3 cos π/4 + sin 2π/3 sin π/4 b. sin 11π/12 = sin(2π/3 + π/4) = sin 2π/3 cos π/4 - cos 2π/3 sin π/4 c. sin 11π/12 = sin(2π/3 + π/4) = sin 2π/3 cos π/4 + cos 2π/3 sin π/4 d. sin 11π/12 = sin(2π/3 - π/4) = cos 2π/3 cos π/4 - sin 2π/3 sin π/4 the exact value of sin 11π/12 is (simplify your answer, including any radicals. use integers or fractions for any numbers)

find the exact value of the following expression. sin 11π/12 rewrite the expression using a sum or difference formula: choose the correct one. a. sin 11π/12 = sin(2π/3 - π/4) = cos 2π/3 cos π/4 + sin 2π/3 sin π/4 b. sin 11π/12 = sin(2π/3 + π/4) = sin 2π/3 cos π/4 - cos 2π/3 sin π/4 c. sin 11π/12 = sin(2π/3 + π/4) = sin 2π/3 cos π/4 + cos 2π/3 sin π/4 d. sin 11π/12 = sin(2π/3 - π/4) = cos 2π/3 cos π/4 - sin 2π/3 sin π/4 the exact value of sin 11π/12 is (simplify your answer, including any radicals. use integers or fractions for any numbers)

Answer

Explanation:

Step1: Recall the sum - of - angles formula for sine

The formula for $\sin(A + B)=\sin A\cos B+\cos A\sin B$. We rewrite $\frac{11\pi}{12}$ as $\frac{2\pi}{3}+\frac{\pi}{4}$ since $\frac{2\pi}{3}+\frac{\pi}{4}=\frac{8\pi + 3\pi}{12}=\frac{11\pi}{12}$. So $\sin\frac{11\pi}{12}=\sin(\frac{2\pi}{3}+\frac{\pi}{4})=\sin\frac{2\pi}{3}\cos\frac{\pi}{4}+\cos\frac{2\pi}{3}\sin\frac{\pi}{4}$.

Step2: Find the values of trigonometric functions

We know that $\sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}$, $\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, $\cos\frac{2\pi}{3}=-\frac{1}{2}$, and $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$. Substitute these values into the formula: [ \begin{align*} \sin\frac{11\pi}{12}&=\sin\frac{2\pi}{3}\cos\frac{\pi}{4}+\cos\frac{2\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:

$\frac{\sqrt{6}-\sqrt{2}}{4}$