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 $$ f(x)=2 $$

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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 $$ f(x) = 2 $$.
There are no solutions for $$ x $$ in the interval $$ [0^{\circ}, 360^{\circ}] $$ that satisfy $$ f(x) = 2 $$.

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