Question

assignment 9.2: sum and differenc
score: 2/8 answered: 2/8
question 3
calculate an exact answer using a formula learned in
\\( \sin 345 ^ { \circ } \\)

Explanation:

Step1: Express \(345^{\circ}\) as a difference

Express \(345^{\circ}\) as \(360^{\circ}-15^{\circ}\). So, \(\sin345^{\circ}=\sin(360^{\circ} - 15^{\circ})\).

Step2: Use the sine difference formula \(\sin(A - B)=\sin A\cos B-\cos A\sin B\)

Here \(A = 360^{\circ}\), \(\sin360^{\circ}=0\), \(\cos360^{\circ}=1\), and \(B = 15^{\circ}\).
\(\sin(360^{\circ}-15^{\circ})=\sin360^{\circ}\cos15^{\circ}-\cos360^{\circ}\sin15^{\circ}\)
Substitute the values: \(0\times\cos15^{\circ}-1\times\sin15^{\circ}=-\sin15^{\circ}\)

Step3: Express \(15^{\circ}\) as a difference \(45^{\circ}-30^{\circ}\)

Use the sine difference formula again: \(\sin(A - B)=\sin A\cos B-\cos A\sin B\), where \(A = 45^{\circ}\), \(B = 30^{\circ}\)
\(\sin15^{\circ}=\sin(45^{\circ}-30^{\circ})=\sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}\)
We know that \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\), \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\), \(\sin30^{\circ}=\frac{1}{2}\)
\(\sin(45^{\circ}-30^{\circ})=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\)

Step4: Find the value of \(\sin345^{\circ}\)

Since \(\sin345^{\circ}=-\sin15^{\circ}\), then \(\sin345^{\circ}=-\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{\sqrt{2}-\sqrt{6}}{4}\)

Answer:

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