Answer :
To prove the identity:
[tex]\[ \frac{\sin^3 A - \cos^3 A}{\sin A - \cos A} = 1 + \frac{1}{2} \sin 2A \][/tex]
we will start by simplifying the left-hand side (LHS) and showing that it equals the right-hand side (RHS).
### Step 1: Simplify the Numerator
The numerator can be simplified using the algebraic identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). Let \(a = \sin A\) and \(b = \cos A\). Then:
[tex]\[ \sin^3 A - \cos^3 A = (\sin A - \cos A) \left((\sin A)^2 + (\sin A)(\cos A) + (\cos A)^2 \right) \][/tex]
### Step 2: Substitute into the LHS
Now substitute back into the LHS:
[tex]\[ \frac{\sin^3 A - \cos^3 A}{\sin A - \cos A} = \frac{(\sin A - \cos A) \left(\sin^2 A + \sin A \cos A + \cos^2 A\right)}{\sin A - \cos A} \][/tex]
The \(\sin A - \cos A\) terms cancel out, leaving:
[tex]\[ \sin^2 A + \sin A \cos A + \cos^2 A \][/tex]
### Step 3: Use the Pythagorean Identity
Recall the Pythagorean identity:
[tex]\[ \sin^2 A + \cos^2 A = 1 \][/tex]
Thus, the expression simplifies to:
[tex]\[ 1 + \sin A \cos A \][/tex]
### Step 4: Express \(\sin A \cos A\) in Terms of \(\sin 2A\)
Recall the double-angle identity for sine:
[tex]\[ \sin 2A = 2 \sin A \cos A \implies \sin A \cos A = \frac{\sin 2A}{2} \][/tex]
Substitute \(\frac{\sin 2A}{2}\) back into our expression:
[tex]\[ 1 + \sin A \cos A = 1 + \frac{\sin 2A}{2} \][/tex]
### Step 5: Compare to the RHS
Notice that this is exactly the form of the RHS:
[tex]\[ 1 + \frac{1}{2} \sin 2A \][/tex]
Thus, we have shown that:
[tex]\[ \frac{\sin^3 A - \cos^3 A}{\sin A - \cos A} = 1 + \frac{1}{2} \sin 2A \][/tex]
### Conclusion
We have proven the given identity:
[tex]\[ \frac{\sin^3 A - \cos^3 A}{\sin A - \cos A} = 1 + \frac{1}{2} \sin 2A \][/tex]
[tex]\[ \frac{\sin^3 A - \cos^3 A}{\sin A - \cos A} = 1 + \frac{1}{2} \sin 2A \][/tex]
we will start by simplifying the left-hand side (LHS) and showing that it equals the right-hand side (RHS).
### Step 1: Simplify the Numerator
The numerator can be simplified using the algebraic identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). Let \(a = \sin A\) and \(b = \cos A\). Then:
[tex]\[ \sin^3 A - \cos^3 A = (\sin A - \cos A) \left((\sin A)^2 + (\sin A)(\cos A) + (\cos A)^2 \right) \][/tex]
### Step 2: Substitute into the LHS
Now substitute back into the LHS:
[tex]\[ \frac{\sin^3 A - \cos^3 A}{\sin A - \cos A} = \frac{(\sin A - \cos A) \left(\sin^2 A + \sin A \cos A + \cos^2 A\right)}{\sin A - \cos A} \][/tex]
The \(\sin A - \cos A\) terms cancel out, leaving:
[tex]\[ \sin^2 A + \sin A \cos A + \cos^2 A \][/tex]
### Step 3: Use the Pythagorean Identity
Recall the Pythagorean identity:
[tex]\[ \sin^2 A + \cos^2 A = 1 \][/tex]
Thus, the expression simplifies to:
[tex]\[ 1 + \sin A \cos A \][/tex]
### Step 4: Express \(\sin A \cos A\) in Terms of \(\sin 2A\)
Recall the double-angle identity for sine:
[tex]\[ \sin 2A = 2 \sin A \cos A \implies \sin A \cos A = \frac{\sin 2A}{2} \][/tex]
Substitute \(\frac{\sin 2A}{2}\) back into our expression:
[tex]\[ 1 + \sin A \cos A = 1 + \frac{\sin 2A}{2} \][/tex]
### Step 5: Compare to the RHS
Notice that this is exactly the form of the RHS:
[tex]\[ 1 + \frac{1}{2} \sin 2A \][/tex]
Thus, we have shown that:
[tex]\[ \frac{\sin^3 A - \cos^3 A}{\sin A - \cos A} = 1 + \frac{1}{2} \sin 2A \][/tex]
### Conclusion
We have proven the given identity:
[tex]\[ \frac{\sin^3 A - \cos^3 A}{\sin A - \cos A} = 1 + \frac{1}{2} \sin 2A \][/tex]
