Question

find the exact value without a calculator. double - angle formulas: $\sin(2\theta)=2\sin\theta\cos\theta$ $\cos(2\theta)=\cos^{2}\theta - \sin^{2}\theta$ $\tan(2\theta)=\frac{2\tan\theta}{1 - \tan^{2}\theta}$ half - angle formulas: $\sin(\frac{\theta}{2})=\pm\sqrt{\frac{1 - \cos\theta}{2}}$ $\cos(\frac{\theta}{2})=\pm\sqrt{\frac{1 + \cos\theta}{2}}$ $\tan(\frac{\theta}{2})=\pm\sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}$ $\sin195^{\circ}=-\frac{\sqrt{?-\sqrt{\square}}}{\square}$

Explanation:

Step1: Rewrite \(195^{\circ}\) as a half - angle

Since \(195^{\circ}=\frac{390^{\circ}}{2}\), and \(\sin195^{\circ}=\sin\frac{390^{\circ}}{2}\). Also, \(195^{\circ}\) is in the second quadrant, so \(\sin195^{\circ}>0\). Using the half - angle formula \(\sin\frac{\theta}{2}=\sqrt{\frac{1 - \cos\theta}{2}}\) with \(\theta = 390^{\circ}\).

Step2: Find \(\cos390^{\circ}\)

We know that \(\cos(360^{\circ}+\alpha)=\cos\alpha\). So, \(\cos390^{\circ}=\cos(360^{\circ} + 30^{\circ})=\cos30^{\circ}=\frac{\sqrt{3}}{2}\).

Step3: Substitute into the half - angle formula

Substitute \(\cos\theta=\frac{\sqrt{3}}{2}\) into \(\sin\frac{\theta}{2}=\sqrt{\frac{1 - \cos\theta}{2}}\). We get \(\sin195^{\circ}=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}\). But we can also use another form. We know that \(\sin(A + B)=\sin A\cos B+\cos A\sin B\), and \(195^{\circ}=150^{\circ}+45^{\circ}\). \(\sin195^{\circ}=\sin(180^{\circ}+15^{\circ})=-\sin15^{\circ}\). And \(\sin15^{\circ}=\sin(45^{\circ}-30^{\circ})=\sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\). So \(\sin195^{\circ}=-\frac{\sqrt{6}-\sqrt{2}}{4}\)

Answer:

\(\sin195^{\circ}=-\frac{\sqrt{6}-\sqrt{2}}{4}\), so the first box is \(6\), the second box is \(2\), and the third box is \(4\)