1) copand \( 0^{\circ} \) b) \( \cos (2 x+y) \) cancuate without usiney a celculator a) \( \sin x \sin y-\cos x \cos y=-\cos 20^{\circ} \) 4 \( \cos 10^{\circ} \mathrm{os} 20^{\circ}-\sin 10^{\circ} \sin 20^{\circ} \) d) \( \cos 62^{\circ} \sin 28^{\circ}-\cos 28^{\circ} \operatorname{in} 242^{\circ} \) e) \( \cos 30^{\circ} \cos 60^{\circ}+\sin 310^{\circ} \sin 200^{\circ} \) Nricosin \( 70^{\circ}-\cos 350 \cos 70^{\circ} \)

Let's solve each part step by step, using trigonometric identities where necessary. ### a) \( \sin x \sin y - \cos x \cos y = -\cos 20^{\circ} \) We can use the cosine of the sum of angles identity: \[ \sin x \sin y - \cos x \cos y = -\cos(x + y) \] Thus, we can rewrite the equation as: \[ -\cos(x + y) = -\cos 20^{\circ} \] This implies: \[ \cos(x + y) = \cos 20^{\circ} \] From this, we can conclude that: \[ x + y = 20^{\circ} + 360^{\circ}k \quad \text{or} \quad x + y = -20^{\circ} + 360^{\circ}k \quad (k \in \mathbb{Z}) \] ### b) \( \cos(2x + y) \) Using the cosine of a sum identity: \[ \cos(2x + y) = \cos(2x)\cos(y) - \sin(2x)\sin(y) \] This expression can be simplified further if we have specific values for \( x \) and \( y \). ### c) \( \cos 10^{\circ} \cos 20^{\circ} - \sin 10^{\circ} \sin 20^{\circ} \) Using the cosine of the sum of angles identity: \[ \cos(10^{\circ} + 20^{\circ}) = \cos 30^{\circ} \] Thus: \[ \cos 10^{\circ} \cos 20^{\circ} - \sin 10^{\circ} \sin 20^{\circ} = \cos 30^{\circ} \] We know that: \[ \cos 30^{\circ} = \frac{\sqrt{3}}{2} \] ### d) \( \cos 62^{\circ} \sin 28^{\circ} - \cos 28^{\circ} \sin 62^{\circ} \) Using the sine of the difference of angles identity: \[ \sin(62^{\circ} - 28^{\circ}) = \sin 34^{\circ} \] Thus: \[ \cos 62^{\circ} \sin 28^{\circ} - \cos 28^{\circ} \sin 62^{\circ} = \sin 34^{\circ} \] ### e) \( \cos 30^{\circ} \cos 60^{\circ} + \sin 310^{\circ} \sin 200^{\circ} \) Using the product-to-sum identities: \[ \cos 30^{\circ} \cos 60^{\circ} = \frac{1}{2} \left( \cos(30^{\circ} - 60^{\circ}) + \cos(30^{\circ} + 60^{\circ}) \right) \] Calculating: \[ \cos 30^{\circ} \cos 60^{\circ} = \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4} \] For the sine terms: \[ \sin 310^{\circ} = -\sin 50^{\circ} \quad \text{and} \quad \sin 200^{\circ} = -\sin 20^{\circ} \] Thus: \[ \sin 310^{\circ} \sin 200^{\circ} = \sin 50^{\circ} \sin 20^{\circ} \] Using the product-to-sum identities again: \[ \sin 50^{\circ} \sin 20^{\circ} = \frac{1}{2} \left( \cos(50^{\circ} - 20^{\circ}) - \cos(50^{\circ} + 20^{\circ}) \right) \] Calculating: \[ \sin 50^{\circ} \sin 20^{\circ} = \frac{1}{2} \left( \cos 30^{\circ} - \cos 70^{\circ} \right) \] ### f) \( \cos 70^{\circ} - \cos 350^{\circ} \cos 70^{\circ} \) Since \( \cos 350^{\circ} = \cos(-10^{\circ}) = \cos 10^{\circ} \): \[ \cos 70^{\circ} - \cos 10^{\circ} \cos 70^{\circ} = \cos 70^{\circ} (1 - \cos 10^{\circ}) \] Now, let's summarize the results: 1. \( x + y = 20^{\circ} + 360^{\circ}k \) or \( x + y = -20^{\circ} + 360^{\circ}k \) 2. \( \cos(2x + y) = \cos(2x)\cos(y) - \sin(2x)\sin(y) \) 3. \( \cos 10^{\circ} \cos 20^{\circ} - \sin 10^{\circ} \sin 20^{\circ} = \frac{\sqrt{3}}{2} \) 4. \( \cos 62^{\circ} \sin 28^{\circ} - \cos 28^{\circ} \sin 62^{\circ} = \sin 34^{\circ} \) 5. \( \cos 30^{\circ} \cos 60^{\circ} + \sin 310^{\circ} \sin 200^{\circ} = \frac{\sqrt{3}}{4} + \frac{1}{2} \left( \cos 30^{\circ} - \cos 70^{\circ} \right) \) 6. \( \cos 70^{\circ} (1 - \cos 10^{\circ}) \) If you need further calculations or simplifications, please let me know!