If `sinA+cosA=m an sin^(3)A+cos^(3)A=n,` then

Options

A. `m^(3)-3m+n=0`
B. `n^(3)-3n+2m=0`
C. `m^(3)-3m+2n=0`
D. `m^(3)+3m+2n=0`

Answer

To solve the problem, we need to derive a relationship between \( m \) and \( n \) based on the given equations. Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Given Equations**:
We have two equations:
\[ \sin A + \cos A = m \]
\[ \sin^3 A + \cos^3 A = n \]

2. **Cubing the First Equation**:
We will cube both sides of the first equation:
\[ (\sin A + \cos A)^3 = m^3 \]

3. **Using the Identity for Cubes**:
We can expand the left-hand side using the identity for the cube of a sum:
\[ a^3 + b^3 + 3ab(a + b) \]
where \( a = \sin A \) and \( b = \cos A \). Thus, we have:
\[ \sin^3 A + \cos^3 A + 3 \sin A \cos A (\sin A + \cos A) = m^3 \]

4. **Substituting Known Values**:
We know that \( \sin^3 A + \cos^3 A = n \) and \( \sin A + \cos A = m \). Substituting these into our expanded equation gives:
\[ n + 3 \sin A \cos A \cdot m = m^3 \]

5. **Using the Identity for Sine and Cosine**:
We can express \( \sin A \cos A \) in terms of \( m \):
\[ \sin A \cos A = \frac{1}{2} \sin(2A) \]
However, we can also use the identity:
\[ \sin^2 A + \cos^2 A = 1 \]
Therefore, we can express \( \sin A \cos A \) as:
\[ \sin A \cos A = \frac{m^2 - 1}{2} \]

6. **Substituting Back**:
Substitute \( \sin A \cos A \) back into the equation:
\[ n + 3 \left(\frac{m^2 - 1}{2}\right) m = m^3 \]

7. **Simplifying the Equation**:
Multiply through by 2 to eliminate the fraction:
\[ 2n + 3m(m^2 - 1) = 2m^3 \]
Expanding gives:
\[ 2n + 3m^3 - 3m = 2m^3 \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ 2n + 3m^3 - 2m^3 - 3m = 0 \]
Simplifying further:
\[ m^3 - 3m + 2n = 0 \]

### Final Result:
The final equation relating \( m \) and \( n \) is:
\[ m^3 - 3m + 2n = 0 \]