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

Let's solve the given problem step by step. We are given:

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

We need to find the relationship between \(m\) and \(n\) that results in one of the given options.

We start by recalling the identity for the sum of cubes:

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

Since \(\sin^2 A + \cos^2 A = 1\), we can write the expression as:

\(= (\sin A + \cos A)(1 - \sin A \cos A)\)

Therefore:

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

Also, recall the identity:

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

 

This means:

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

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

Substitute back:

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

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

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

\(n = m \cdot \frac{3 - m^2}{2}\)

Therefore, \(2n = m(3 - m^2)\).

Rearranging gives:

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

Rewriting this expression, we get:

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

Hence, the correct answer is:

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