If `cos(3x-20^(@))=sin(3y+20^(@))` then the value of `(x+y)` is

Options

A. `20^(@)`
B. `30^(@)`
C. `40^(@)`
D. `45^(@)`

Answer

To solve the equation \( \cos(3x - 20^\circ) = \sin(3y + 20^\circ) \), we can use the identity that relates sine and cosine. Specifically, we know that:

\[ \sin \theta = \cos(90^\circ - \theta) \]

### Step 1: Rewrite the sine function
Using the identity, we can rewrite \( \sin(3y + 20^\circ) \):

\[ \sin(3y + 20^\circ) = \cos(90^\circ - (3y + 20^\circ)) = \cos(90^\circ - 3y - 20^\circ) = \cos(70^\circ - 3y) \]

### Step 2: Set the cosines equal
Now we can set the two cosine expressions equal to each other:

\[ \cos(3x - 20^\circ) = \cos(70^\circ - 3y) \]

### Step 3: Use the cosine identity
The cosine function has the property that \( \cos A = \cos B \) implies:

\[ A = B + 360^\circ n \quad \text{or} \quad A = -B + 360^\circ n \]

for some integer \( n \). Therefore, we can write two equations:

1. \( 3x - 20^\circ = 70^\circ - 3y + 360^\circ n \)
2. \( 3x - 20^\circ = - (70^\circ - 3y) + 360^\circ n \)

### Step 4: Solve the first equation
Let's solve the first equation:

\[ 3x - 20^\circ = 70^\circ - 3y + 360^\circ n \]

Rearranging gives:

\[ 3x + 3y = 90^\circ + 360^\circ n \]

Dividing everything by 3:

\[ x + y = 30^\circ + 120^\circ n \]

### Step 5: Solve the second equation
Now, let's solve the second equation:

\[ 3x - 20^\circ = -70^\circ + 3y + 360^\circ n \]

Rearranging gives:

\[ 3x - 3y = -50^\circ + 360^\circ n \]

Dividing everything by 3:

\[ x - y = -\frac{50^\circ}{3} + 120^\circ n \]

### Step 6: Solve the system of equations
Now we have two equations:

1. \( x + y = 30^\circ + 120^\circ n \)
2. \( x - y = -\frac{50^\circ}{3} + 120^\circ n \)

We can add these two equations:

\[ (x + y) + (x - y) = (30^\circ + 120^\circ n) + \left(-\frac{50^\circ}{3} + 120^\circ n\right) \]

This simplifies to:

\[ 2x = 30^\circ - \frac{50^\circ}{3} + 240^\circ n \]

Calculating \( 30^\circ - \frac{50^\circ}{3} \):

\[ 30^\circ = \frac{90^\circ}{3} \quad \Rightarrow \quad 30^\circ - \frac{50^\circ}{3} = \frac{90^\circ - 50^\circ}{3} = \frac{40^\circ}{3} \]

Thus, we have:

\[ 2x = \frac{40^\circ}{3} + 240^\circ n \]

Dividing by 2:

\[ x = \frac{20^\circ}{3} + 120^\circ n \]

Now substituting \( x \) back into the first equation to find \( y \):

\[ \frac{20^\circ}{3} + 120^\circ n + y = 30^\circ + 120^\circ n \]

This simplifies to:

\[ y = 30^\circ - \frac{20^\circ}{3} = \frac{90^\circ}{3} - \frac{20^\circ}{3} = \frac{70^\circ}{3} \]

### Step 7: Find \( x + y \)
Now we can find \( x + y \):

\[ x + y = \left(\frac{20^\circ}{3} + 120^\circ n\right) + \left(\frac{70^\circ}{3}\right) = \frac{90^\circ}{3} + 120^\circ n = 30^\circ + 120^\circ n \]

Thus, the value of \( x + y \) is:

\[ \boxed{30^\circ} \]