Gomework Given identiey \( \cos (x-1)=\cos x \cdot \cos y+\sin x \cdot \sin y \) bi Compound angle identity given above the derieve the formula \( \cos (x+y) \) 102 Hences or otherwi.st show that \( \frac{\cos \left(90^{\circ}-x\right) \cdot \cos y+\sin (-y) \cdot \cos (180+x)}{\cos x \cdot \cos (360+y)+\sin \left(360^{\circ}-x\right) \cdot \sin y} \) Classwork Given \( f(x)=\sqrt{6 \sin ^{2} x-11 \cos (90+x)+7} \) solve for \( x \) in the interv \( x \in\left[0^{\circ}, 360\right] \) if \( f(x)=2 \)

Let's break down the problem step by step. ### Part 1: Deriving the formula for \( \cos(x+y) \) We start with the given identity: \[ \cos(x-1) = \cos x \cdot \cos y + \sin x \cdot \sin y \] This is a form of the cosine of a difference identity. To derive the formula for \( \cos(x+y) \), we can use the cosine of a sum identity, which states: \[ \cos(x+y) = \cos x \cdot \cos y - \sin x \cdot \sin y \] To show that the two identities are related, we can manipulate the first identity. 1. **Rearranging the first identity**: \[ \cos(x-1) = \cos x \cdot \cos y + \sin x \cdot \sin y \] We can express \( \cos(x-1) \) using the cosine of a sum: \[ \cos(x-1) = \cos x \cdot \cos 1 + \sin x \cdot \sin 1 \] 2. **Setting \( y = 1 \)**: If we set \( y = 1 \), we can see that: \[ \cos(x-1) = \cos x \cdot \cos 1 + \sin x \cdot \sin 1 \] This matches the form of the cosine of a sum identity. 3. **Conclusion**: Thus, we can conclude that: \[ \cos(x+y) = \cos x \cdot \cos y - \sin x \cdot \sin y \] ### Part 2: Simplifying the expression Now we need to simplify the expression: \[ \frac{\cos(90^{\circ}-x) \cdot \cos y + \sin(-y) \cdot \cos(180+x)}{\cos x \cdot \cos(360+y) + \sin(360^{\circ}-x) \cdot \sin y} \] 1. **Using trigonometric identities**: - \( \cos(90^{\circ}-x) = \sin x \) - \( \sin(-y) = -\sin y \) - \( \cos(180+x) = -\cos x \) - \( \cos(360+y) = \cos y \) - \( \sin(360^{\circ}-x) = -\sin x \) 2. **Substituting these identities**: The numerator becomes: \[ \sin x \cdot \cos y - \sin y \cdot (-\cos x) = \sin x \cdot \cos y + \sin y \cdot \cos x \] The denominator becomes: \[ \cos x \cdot \cos y - \sin x \cdot \sin y \] 3. **Final expression**: Thus, the expression simplifies to: \[ \frac{\sin x \cdot \cos y + \sin y \cdot \cos x}{\cos x \cdot \cos y - \sin x \cdot \sin y} \] ### Part 3: Solving for \( x \) in \( f(x) = 2 \) Given: \[ f(x) = \sqrt{6 \sin^2 x - 11 \cos(90+x) + 7} \] We need to solve for \( x \) in the interval \( [0^{\circ}, 360^{\circ}] \) when \( f(x) = 2 \). 1. **Setting up the equation**: \[ \sqrt{6 \sin^2 x - 11 \cos(90+x) + 7} = 2 \] 2. **Squaring both sides**: \[ 6 \sin^2 x - 11 \cos(90+x) + 7 = 4 \] Simplifying gives: \[ 6 \sin^2 x - 11(-\sin x) + 7 = 4 \] \[ 6 \sin^2 x + 11 \sin x + 3 = 0 \] 3. **Letting \( y = \sin x \)**: The equation becomes: \[ 6y^2 + 11y + 3 = 0 \] 4. **Using the quadratic formula**: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 6 \cdot 3}}{2 \cdot 6} \] \[ = \frac{-11 \pm \sqrt{121 - 72}}{12} = \frac{-11 \pm \sqrt{49}}{12} = \frac{-11 \pm 7}{12} \] 5. **Calculating the roots**: - \( y_1 = \frac{-4}{12} = -\frac{1}{3} \) (not valid since \( \sin x \) must be between -1 and 1) - \( y_2 = \frac{-18}{12} = -\frac{3}{2} \) (not valid) Since both roots are invalid, we need to check if there are any solutions in the interval \( [0^{\circ}, 360^{\circ}] \) for \( f(x) = 2 \). ### Conclusion The expression simplifies to a form that can be analyzed further, and the function \( f(x) = 2 \) does not yield valid solutions in the specified interval. Thus, there are no solutions for \( x \) in \( [0^{\circ}, 360^{\circ}] \) such that \( f(x) = 2 \).