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 establish a relationship between \( m \) and \( n \) given the equations:

1. \( \sin A + \cos A = m \)
2. \( \sin^3 A + \cos^3 A = n \)

We will use the identity for the sum of cubes and some trigonometric identities to derive the required relationship.

### Step-by-Step Solution:

**Step 1: Use the identity for the sum of cubes.**

The identity for the sum of cubes states:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
In our case, let \( a = \sin A \) and \( b = \cos A \). Therefore,
\[ \sin^3 A + \cos^3 A = (\sin A + \cos A)(\sin^2 A - \sin A \cos A + \cos^2 A) \]

**Step 2: Substitute \( \sin A + \cos A \) with \( m \).**

From the first equation, we know that:
\[ \sin A + \cos A = m \]
Now, we need to find \( \sin^2 A + \cos^2 A \) and \( \sin A \cos A \).

Using the Pythagorean identity:
\[ \sin^2 A + \cos^2 A = 1 \]

**Step 3: Express \( \sin^2 A - \sin A \cos A + \cos^2 A \).**

Now we can substitute into our equation:
\[ \sin^3 A + \cos^3 A = m(1 - \sin A \cos A) \]

**Step 4: Express \( \sin A \cos A \) in terms of \( m \).**

Using the identity:
\[ \sin A \cos A = \frac{1}{2} \sin(2A) \]
We can also express \( \sin A \cos A \) in terms of \( m \):
\[ \sin A \cos A = \frac{m^2 - 1}{2} \]
This comes from the expansion of \( (\sin A + \cos A)^2 \):
\[ m^2 = \sin^2 A + \cos^2 A + 2\sin A \cos A = 1 + 2\sin A \cos A \]
Thus,
\[ 2\sin A \cos A = m^2 - 1 \implies \sin A \cos A = \frac{m^2 - 1}{2} \]

**Step 5: Substitute back into the equation for \( n \).**

Now substituting \( \sin A \cos A \) back into the equation for \( n \):
\[ n = m \left(1 - \frac{m^2 - 1}{2}\right) \]
This simplifies to:
\[ n = m \left(1 - \frac{m^2}{2} + \frac{1}{2}\right) = m \left(\frac{3}{2} - \frac{m^2}{2}\right) \]
Thus,
\[ n = \frac{m(3 - m^2)}{2} \]

**Step 6: Rearranging the equation.**

To find a polynomial equation, we can multiply both sides by 2:
\[ 2n = m(3 - m^2) \]
Rearranging gives:
\[ m^3 - 3m + 2n = 0 \]

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