determine the exact value for each trigonometric expression. if you have a fraction in your answer like…

determine the exact value for each trigonometric expression. if you have a fraction in your answer like this, \\( \\frac { \\sqrt { 3 } - \\sqrt { 2 } } { 2 } \\), answer it like this, (sqrt(3)-sqrt(2))/2. make sure you have the parentheses around the numerator. \\( \\cos \\left( - \\frac { \\pi } { 12 } \\right) \\) answer: determine the exact value for each trigonometric expression. if you have a fraction in your answer like this, \\( \\frac { \\sqrt { 3 } - \\sqrt { 2 } } { 2 } \\), answer it like this, (sqrt(3)-sqrt(2))/2. make sure you have the parentheses around the numerator. \\( \\cos 165 ^ { \\circ } \\) answer:

determine the exact value for each trigonometric expression. if you have a fraction in your answer like this, \\( \\frac { \\sqrt { 3 } - \\sqrt { 2 } } { 2 } \\), answer it like this, (sqrt(3)-sqrt(2))/2. make sure you have the parentheses around the numerator. \\( \\cos \\left( - \\frac { \\pi } { 12 } \\right) \\) answer: determine the exact value for each trigonometric expression. if you have a fraction in your answer like this, \\( \\frac { \\sqrt { 3 } - \\sqrt { 2 } } { 2 } \\), answer it like this, (sqrt(3)-sqrt(2))/2. make sure you have the parentheses around the numerator. \\( \\cos 165 ^ { \\circ } \\) answer:

Answer

Explanation:

Step1: Use the cosine of a difference formula for (\cos(-\frac{\pi}{12}))

We know that (\cos(A - B)=\cos A\cos B+\sin A\sin B) and (\cos(-x)=\cos x). Also, (\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}). So, (\cos(\frac{\pi}{12})=\cos(\frac{\pi}{3}-\frac{\pi}{4})=\cos\frac{\pi}{3}\cos\frac{\pi}{4}+\sin\frac{\pi}{3}\sin\frac{\pi}{4}) Since (\cos\frac{\pi}{3}=\frac{1}{2}), (\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}), (\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}), (\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}) [ \begin{align*} \cos(\frac{\pi}{12})&=\frac{1}{2}\times\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}\ &=\frac{\sqrt{2}(1 + \sqrt{3})}{4} \end{align*} ]

Step2: Use the cosine of a sum formula for (\cos165^{\circ})

We know that (\cos(A + B)=\cos A\cos B-\sin A\sin B) and (165^{\circ}=120^{\circ}+45^{\circ}) (\cos165^{\circ}=\cos(120^{\circ}+45^{\circ})=\cos120^{\circ}\cos45^{\circ}-\sin120^{\circ}\sin45^{\circ}) Since (\cos120^{\circ}=-\frac{1}{2}), (\cos45^{\circ}=\frac{\sqrt{2}}{2}), (\sin120^{\circ}=\frac{\sqrt{3}}{2}), (\sin45^{\circ}=\frac{\sqrt{2}}{2}) [ \begin{align*} \cos165^{\circ}&=-\frac{1}{2}\times\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}\ &=-\frac{\sqrt{2}(1+\sqrt{3})}{4} \end{align*} ]

Answer:

For (\cos(-\frac{\pi}{12})): ((sqrt(2)+sqrt(6))/4) For (\cos165^{\circ}): (-(sqrt(2)+sqrt(6))/4)