Simplify fully: \( \sin \left(90^{\circ}-x\right) \cdot \cos \left(180^{\circ}+x\right)+\tan x \cdot \cos x \cdot \sin \left(x-180^{\circ}\right) \) Prove, WITHOUT using a calculator, that \( \frac{\sin 315^{\circ} \cdot \tan 210^{\circ} \cdot \sin 190^{\circ}}{\cos 100^{\circ} \cdot \sin 120^{\circ}}=\frac{-\sqrt{2}}{3} \)
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To simplify the expression \( \sin(90^{\circ} - x) \cdot \cos(180^{\circ} + x) + \tan x \cdot \cos x \cdot \sin(x - 180^{\circ}) \): 1. We know that \( \sin(90^{\circ} - x) = \cos x \). 2. Also, \( \cos(180^{\circ} + x) = -\cos x \). 3. Therefore, \( \sin(90^{\circ} - x) \cdot \cos(180^{\circ} + x) = \cos x \cdot (-\cos x) = -\cos^2 x \). Next, simplify the second part: 1. Recall that \( \tan x = \frac{\sin x}{\cos x} \) and \( \sin(x - 180^{\circ}) = -\sin x \). 2. Thus, \( \tan x \cdot \cos x \cdot \sin(x - 180^{\circ}) = \frac{\sin x}{\cos x} \cdot \cos x \cdot (-\sin x) = -\sin^2 x \). Combining both parts, we have: \[ -\cos^2 x - \sin^2 x = -(\sin^2 x + \cos^2 x) = -1. \] Thus, the full simplification results in \( -1 \). --- Next, for proving \( \frac{\sin 315^{\circ} \cdot \tan 210^{\circ} \cdot \sin 190^{\circ}}{\cos 100^{\circ} \cdot \sin 120^{\circ}} = \frac{-\sqrt{2}}{3} \): 1. First, calculate \( \sin 315^{\circ} = \sin(360^{\circ} - 45^{\circ}) = -\sin 45^{\circ} = -\frac{\sqrt{2}}{2} \). 2. Next, \( \tan 210^{\circ} = \tan(180^{\circ} + 30^{\circ}) = \tan 30^{\circ} = \frac{1}{\sqrt{3}} \). 3. Then, \( \sin 190^{\circ} = \sin(180^{\circ} + 10^{\circ}) = -\sin 10^{\circ} \). For the denominator: 1. We have \( \cos 100^{\circ} = -\sin 10^{\circ} \) and \( \sin 120^{\circ} = \sin(180^{\circ} - 60^{\circ}) = \sin 60^{\circ} = \frac{\sqrt{3}}{2} \). Putting it all together: \[ \frac{\left(-\frac{\sqrt{2}}{2}\right) \cdot \left(\frac{1}{\sqrt{3}}\right) \cdot \left(-\sin 10^{\circ}\right)}{-\sin 10^{\circ} \cdot \frac{\sqrt{3}}{2}}. \] Simplifying leads to: \[ \frac{\frac{\sqrt{2}}{2} \cdot \frac{1}{\sqrt{3}}}{-\frac{\sqrt{3}}{2}} = \frac{\sqrt{2}}{2} \cdot \frac{2}{\sqrt{3}} \cdot \frac{1}{\frac{-\sqrt{3}}{2}} = \frac{\sqrt{2}}{3}, \] thus concluding the left side is indeed \( \frac{-\sqrt{2}}{3} \).