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 \( \frac{\sin 315^{\circ} \cdot \tan 210^{\circ} \cdot \sin 190^{\circ}}{\cos 100^{\circ} \cdot \sin 120^{\circ}}=\frac{-\sqrt{2}}{3} \)

Alright, I need to simplify the given trigonometric expression and then prove the identity without using a calculator. Let's start with the simplification. The expression is: \[ \sin \left(90^{\circ}-x\right) \cdot \cos \left(180^{\circ}+x\right) + \tan x \cdot \cos x \cdot \sin \left(x-180^{\circ}\right) \] First, I recall some trigonometric identities that might help: 1. \(\sin(90^{\circ} - x) = \cos x\) 2. \(\cos(180^{\circ} + x) = -\cos x\) 3. \(\tan x = \frac{\sin x}{\cos x}\) 4. \(\sin(x - 180^{\circ}) = -\sin x\) Applying these identities to the expression: \[ \cos x \cdot (-\cos x) + \frac{\sin x}{\cos x} \cdot \cos x \cdot (-\sin x) \] Simplifying each term: 1. \(\cos x \cdot (-\cos x) = -\cos^2 x\) 2. \(\frac{\sin x}{\cos x} \cdot \cos x \cdot (-\sin x) = \sin x \cdot (-\sin x) = -\sin^2 x\) So the expression becomes: \[ -\cos^2 x - \sin^2 x \] I know that \(\cos^2 x + \sin^2 x = 1\), so: \[ -(\cos^2 x + \sin^2 x) = -1 \] Therefore, the simplified form of the expression is \(-1\). Now, moving on to the proof: \[ \frac{\sin 315^{\circ} \cdot \tan 210^{\circ} \cdot \sin 190^{\circ}}{\cos 100^{\circ} \cdot \sin 120^{\circ}} = \frac{-\sqrt{2}}{3} \] First, let's evaluate each trigonometric function: 1. \(\sin 315^{\circ}\): - \(315^{\circ}\) is in the fourth quadrant where sine is negative. - \(315^{\circ} = 360^{\circ} - 45^{\circ}\), so \(\sin 315^{\circ} = -\sin 45^{\circ} = -\frac{\sqrt{2}}{2}\) 2. \(\tan 210^{\circ}\): - \(210^{\circ}\) is in the third quadrant where tangent is positive. - \(210^{\circ} = 180^{\circ} + 30^{\circ}\), so \(\tan 210^{\circ} = \tan 30^{\circ} = \frac{\sqrt{3}}{3}\) 3. \(\sin 190^{\circ}\): - \(190^{\circ}\) is in the third quadrant where sine is negative. - \(190^{\circ} = 180^{\circ} + 10^{\circ}\), so \(\sin 190^{\circ} = -\sin 10^{\circ}\) 4. \(\cos 100^{\circ}\): - \(100^{\circ}\) is in the second quadrant where cosine is negative. - \(\cos 100^{\circ} = -\cos 80^{\circ}\) 5. \(\sin 120^{\circ}\): - \(120^{\circ}\) is in the second quadrant where sine is positive. - \(\sin 120^{\circ} = \sin 60^{\circ} = \frac{\sqrt{3}}{2}\) Now, plug these values back into the original expression: \[ \frac{(-\frac{\sqrt{2}}{2}) \cdot (\frac{\sqrt{3}}{3}) \cdot (-\sin 10^{\circ})}{(-\cos 80^{\circ}) \cdot (\frac{\sqrt{3}}{2})} \] Simplify the numerator and the denominator separately: Numerator: \[ (-\frac{\sqrt{2}}{2}) \cdot (\frac{\sqrt{3}}{3}) \cdot (-\sin 10^{\circ}) = \frac{\sqrt{2} \cdot \sqrt{3} \cdot \sin 10^{\circ}}{6} \] Denominator: \[ (-\cos 80^{\circ}) \cdot (\frac{\sqrt{3}}{2}) = -\frac{\sqrt{3} \cdot \cos 80^{\circ}}{2} \] So the expression becomes: \[ \frac{\frac{\sqrt{2} \cdot \sqrt{3} \cdot \sin 10^{\circ}}{6}}{-\frac{\sqrt{3} \cdot \cos 80^{\circ}}{2}} = \frac{\sqrt{2} \cdot \sqrt{3} \cdot \sin 10^{\circ} \cdot 2}{6 \cdot \sqrt{3} \cdot \cos 80^{\circ}} = \frac{\sqrt{2} \cdot \sin 10^{\circ}}{3 \cdot \cos 80^{\circ}} \] Now, recall that \(\cos 80^{\circ} = \sin 10^{\circ}\) because \(\cos(90^{\circ} - \theta) = \sin \theta\). Therefore: \[ \cos 80^{\circ} = \sin 10^{\circ} \] Substituting back: \[ \frac{\sqrt{2} \cdot \sin 10^{\circ}}{3 \cdot \sin 10^{\circ}} = \frac{\sqrt{2}}{3} \] However, the original expression is supposed to equal \(-\frac{\sqrt{2}}{3}\). This discrepancy suggests that there might be a negative sign missing or an error in the initial assumption. Upon reviewing, I realize that the negative signs in the numerator and denominator should cancel out, leading to a positive \(\frac{\sqrt{2}}{3}\). Therefore, the correct simplified form should be \(\frac{\sqrt{2}}{3}\), not \(-\frac{\sqrt{2}}{3}\).