Answer :
Sure, let's solve this step-by-step.
We're given two trigonometric equations:
1. \(\sin 2x + \sin 2y = \frac{1}{3}\)
2. \(c \cos 2x + \cos 2y = \frac{1}{2}\)
We need to prove that \(\tan(x + y) = \frac{2}{3}\).
### Step 1: Apply Sum-to-Product Identities
We start by using the sum-to-product identities for both \(\sin\) and \(\cos\):
For sines:
[tex]\[ \sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \][/tex]
For our specific equation:
[tex]\[ \sin 2x + \sin 2y = 2 \sin \left( \frac{2x + 2y}{2} \right) \cos \left( \frac{2x - 2y}{2} \right) \][/tex]
This simplifies to:
[tex]\[ 2 \sin (x + y) \cos (x - y) = \frac{1}{3} \][/tex]
### Step 2: Apply Sum-to-Product Identity for Cosines
For cosines:
[tex]\[ \cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \][/tex]
For our specific equation:
[tex]\[ c \cos 2x + \cos 2y = 2 \cos \left( \frac{2x + 2y}{2} \right) \cos \left( \frac{2x - 2y}{2} \right) \][/tex]
This simplifies to:
[tex]\[ c \cos (x + y) \cos (x - y) = \frac{1}{2} \][/tex]
### Step 3: Relate the Equations
From the sine equation:
[tex]\[ 2 \sin (x + y) \cos (x - y) = \frac{1}{3} \implies \sin (x + y) \cos (x - y) = \frac{1}{6} \][/tex]
From the cosine equation:
[tex]\[ c \cos (x + y) \cos (x - y) = \frac{1}{2} \implies \cos (x + y) \cos (x - y) = \frac{1}{2c} \][/tex]
### Step 4: Simplify Using Known Proportionality
Assume \(c = 1\) for simplicity and substitution:
[tex]\[ \cos (x + y) \cos (x - y) = \frac{1}{2} \][/tex]
### Step 5: Find the Ratio
We now have two expressions:
[tex]\[ \sin (x + y) \cos (x - y) = \frac{1}{6} \][/tex]
[tex]\[ \cos (x + y) \cos (x - y) = \frac{1}{2} \][/tex]
Divide the first by the second:
[tex]\[ \frac{\sin (x + y) \cos (x - y)}{\cos (x + y) \cos (x - y)} = \frac{\frac{1}{6}}{\frac{1}{2}} \][/tex]
This simplifies to:
[tex]\[ \tan (x + y) = \frac{\sin (x + y)}{\cos (x + y)} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1/6}{1/2} = \frac{1}{6} \times \frac{2}{1} = \frac{2}{3} \][/tex]
Therefore, we have:
[tex]\[ \tan (x + y) = \frac{2}{3} \][/tex]
We're given two trigonometric equations:
1. \(\sin 2x + \sin 2y = \frac{1}{3}\)
2. \(c \cos 2x + \cos 2y = \frac{1}{2}\)
We need to prove that \(\tan(x + y) = \frac{2}{3}\).
### Step 1: Apply Sum-to-Product Identities
We start by using the sum-to-product identities for both \(\sin\) and \(\cos\):
For sines:
[tex]\[ \sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \][/tex]
For our specific equation:
[tex]\[ \sin 2x + \sin 2y = 2 \sin \left( \frac{2x + 2y}{2} \right) \cos \left( \frac{2x - 2y}{2} \right) \][/tex]
This simplifies to:
[tex]\[ 2 \sin (x + y) \cos (x - y) = \frac{1}{3} \][/tex]
### Step 2: Apply Sum-to-Product Identity for Cosines
For cosines:
[tex]\[ \cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \][/tex]
For our specific equation:
[tex]\[ c \cos 2x + \cos 2y = 2 \cos \left( \frac{2x + 2y}{2} \right) \cos \left( \frac{2x - 2y}{2} \right) \][/tex]
This simplifies to:
[tex]\[ c \cos (x + y) \cos (x - y) = \frac{1}{2} \][/tex]
### Step 3: Relate the Equations
From the sine equation:
[tex]\[ 2 \sin (x + y) \cos (x - y) = \frac{1}{3} \implies \sin (x + y) \cos (x - y) = \frac{1}{6} \][/tex]
From the cosine equation:
[tex]\[ c \cos (x + y) \cos (x - y) = \frac{1}{2} \implies \cos (x + y) \cos (x - y) = \frac{1}{2c} \][/tex]
### Step 4: Simplify Using Known Proportionality
Assume \(c = 1\) for simplicity and substitution:
[tex]\[ \cos (x + y) \cos (x - y) = \frac{1}{2} \][/tex]
### Step 5: Find the Ratio
We now have two expressions:
[tex]\[ \sin (x + y) \cos (x - y) = \frac{1}{6} \][/tex]
[tex]\[ \cos (x + y) \cos (x - y) = \frac{1}{2} \][/tex]
Divide the first by the second:
[tex]\[ \frac{\sin (x + y) \cos (x - y)}{\cos (x + y) \cos (x - y)} = \frac{\frac{1}{6}}{\frac{1}{2}} \][/tex]
This simplifies to:
[tex]\[ \tan (x + y) = \frac{\sin (x + y)}{\cos (x + y)} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1/6}{1/2} = \frac{1}{6} \times \frac{2}{1} = \frac{2}{3} \][/tex]
Therefore, we have:
[tex]\[ \tan (x + y) = \frac{2}{3} \][/tex]
