Question

If \(\sin A + \cos A = m\) and \(\sin^3 A + \cos^3 A = n\), then

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

Hint:

\(\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)\).

Answer 1

The Correct Option is C

Given two equations: 

• \(\sin A + \cos A = m\)
• \(\sin^3 A + \cos^3 A = n\)

We need to find the relationship between \(m\) and \(n\). Let's begin by using an algebraic identity for the sum of cubes:

The identity for the sum of cubes is:

\(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)

By letting \(a = \sin A\) and \(b = \cos A\), we use this identity to express:

\(\sin^3 A + \cos^3 A = (\sin A + \cos A)((\sin A)^2 - \sin A \cos A + (\cos A)^2)\)

Now substitute the given \(\sin A + \cos A = m\):

\(\sin^3 A + \cos^3 A = m((\sin A)^2 - \sin A \cos A + (\cos A)^2)\)

Note that \((\sin A)^2 + (\cos A)^2 = 1\) (Pythagorean identity). Hence:

\((\sin A)^2 + (\cos A)^2 = 1 \implies (\sin A)^2 + (\cos A)^2 = 1 - \sin A \cos A\)

Therefore, \(\sin A^2 - \sin A \cos A + \cos A^2 = 1 - \sin A \cos A\) can be rewritten as:\)

\(\sin^3 A + \cos^3 A = m(1 - \sin A \cos A)\)

Given that \(n = \sin^3 A + \cos^3 A\), we substitute:

\(n = m(1 - \sin A \cos A)\)

From the square of the first equation, we have:

\((\sin A + \cos A)^2 = m^2 = \sin^2 A + \cos^2 A + 2\sin A \cos A\)\)

Using \(\sin^2 A + \cos^2 A = 1\), we find:

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

Solving for \(\sin A \cos A\) gives:

\(\sin A \cos A = \frac{m^2 - 1}{2}\)

Substitute this back into the expression for \(n\):

\(n = m\left(1 - \frac{m^2 - 1}{2}\right)\)

Simplify this expression:

\(n = m\left(1 - \frac{m^2}{2} + \frac{1}{2}\right) = m \left(\frac{2 - m^2 + 1}{2}\right) = m\left(\frac{3 - m^2}{2}\right) = \frac{m(3 - m^2)}{2}\)

This simplifies to: \(2n = m(3 - m^2)\)\)

Expanding this gives:

\(2n = 3m - m^3\)

Rewriting yields the equation:

\(m^3 - 3m + 2n = 0\)

Thus, the correct answer is:

• \(m^3 - 3m + 2n = 0\)