b) $$ \cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2} $$

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UpStudy AI · Accepted Answer

Certainly! Let's delve into proving the trigonometric identity:

$$
\cos x + \cos y = 2 \cos \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)
$$

### **Proof Using Sum-to-Product Formulas**

The identity you're looking to prove is one of the standard **sum-to-product** formulas in trigonometry. These formulas convert sums or differences of trigonometric functions into products, which can simplify many proofs and calculations.

#### **Step 1: Recall the Sum-to-Product Formula**

One of the sum-to-product identities is:

$$
\cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)
$$

In this identity, $$ A $$ and $$ B $$ are any angles. Comparing this with the identity you provided:

$$
\cos x + \cos y = 2 \cos \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)
$$

We can see that it's a direct application of the sum-to-product formula where $$ A = x $$ and $$ B = y $$.

#### **Step 2: Derive the Identity Using Exponential Form (Euler's Formula)**

Alternatively, we can derive the identity using Euler's formula, which relates trigonometric functions to complex exponentials:

$$
\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}
$$

Let's apply this to $$ \cos x + \cos y $$:

$$
\cos x + \cos y = \frac{e^{ix} + e^{-ix}}{2} + \frac{e^{iy} + e^{-iy}}{2} = \frac{e^{ix} + e^{-ix} + e^{iy} + e^{-iy}}{2}
$$

Now, let's factor the terms:

$$
\frac{e^{i(x+y)/2} \left( e^{i(x-y)/2} + e^{-i(x-y)/2} \right) + e^{-i(x+y)/2} \left( e^{i(x-y)/2} + e^{-i(x-y)/2} \right)}{2}
$$

Recognizing that $$ e^{i\theta} + e^{-i\theta} = 2\cos \theta $$, we can simplify:

$$
= \frac{2 \cos \left( \frac{x - y}{2} \right) \left[ e^{i(x+y)/2} + e^{-i(x+y)/2} \right]}{2}
$$

Again, using $$ e^{i\theta} + e^{-i\theta} = 2\cos \theta $$:

$$
= 2 \cos \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)
$$

Hence, we've derived:

$$
\cos x + \cos y = 2 \cos \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)
$$

### **Conclusion**

The identity is proven using the sum-to-product formula and can also be derived through Euler's formula. This identity is particularly useful in simplifying expressions involving the sum of cosines or in solving trigonometric equations.
$$ \cos x + \cos y = 2 \cos \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right) $$

b) \( \cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2} \)

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