Solving for $\sin^3 A + \cos^3 A$

We are given the equation: $ \sin A + \cos A = \sqrt{2} $ Our goal is to find the value of $\sin^3 A + \cos^3 A$.

Deriving $\sin A \cos A$

Square both sides of the given equation:

$ (\sin A + \cos A)^2 = (\sqrt{2})^2 $

Expand the left side:

$ \sin^2 A + \cos^2 A + 2 \sin A \cos A = 2 $

Using the identity $\sin^2 A + \cos^2 A = 1$, we get:

$ 1 + 2 \sin A \cos A = 2 $

Solve for $2 \sin A \cos A$:

$ 2 \sin A \cos A = 2 - 1 $

$ 2 \sin A \cos A = 1 $

Therefore:

$ \sin A \cos A = \frac{1}{2} $

Calculating $\sin^3 A + \cos^3 A$

We use the algebraic identity for the sum of cubes: $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$.

Let $x = \sin A$ and $y = \cos A$. Applying the identity:

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

Rearrange the terms in the second factor:

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

Now substitute the known values:

• $\sin A + \cos A = \sqrt{2}$
• $\sin^2 A + \cos^2 A = 1$
• $\sin A \cos A = \frac{1}{2}$

Substitute these values into the equation:

$ \sin^3 A + \cos^3 A = (\sqrt{2}) \left( 1 - \frac{1}{2} \right) $

$ \sin^3 A + \cos^3 A = (\sqrt{2}) \left( \frac{1}{2} \right) $

$ \sin^3 A + \cos^3 A = \frac{\sqrt{2}}{2} $

The value of $\sin^3 A + \cos^3 A$ is $\frac{\sqrt{2}}{2}$.