Trigonometry & Inverse Trigonometry
Symmetric Functions
Grade 11
Question:
<p>If \(x = \sin(\alpha - \beta) \times \sin(\gamma - \delta)\), \(y = \sin(\beta - \gamma) \times \sin(\alpha - \delta)\), \(z = \sin(\gamma - \alpha) \times \sin(\beta - \delta)\), then :</p>
<p>(a) \(x + y + z = 0\)</p>
<p>(b) \(x^3 + y^3 + z^3 = 3xyz\)</p>
<p>(c) \(x + y - z = 0\)</p>
<p>(d) \(x^3 + y^3 - z^3 = 3xyz\)</p>
Step-by-Step Solution
Key Concept: Use the product-to-sum trigonometric identity and algebraic manipulation to establish a relationship between x, y, z. Recognize that these expressions satisfy both x + y + z = 0 and the cubic identity x³ + y³ + z³ = 3xyz simultaneously.
<p><strong>Step 1: Apply the Product-to-Sum Identity</strong></p><p>Use the identity: sin(A)sin(B) = ½[cos(A-B) - cos(A+B)]</p><p>For each term:</p><p>x = sin(α-β)sin(γ-δ) = ½[cos((α-β)-(γ-δ)) - cos((α-β)+(γ-δ))]</p><p>x = ½[cos(α-β-γ+δ) - cos(α-β+γ-δ)]</p><p><strong>Step 2: Use the Constraint That x + y + z Must Be Evaluated</strong></p><p>Similarly compute y and z:</p><p>y = sin(β-γ)sin(α-δ) = ½[cos(β-γ-α+δ) - cos(β-γ+α-δ)]</p><p>z = sin(γ-α)sin(β-δ) = ½[cos(γ-α-β+δ) - cos(γ-α+β-δ)]</p><p><strong>Step 3: Add x + y + z</strong></p><p>Notice that:</p><p>• cos(α-β-γ+δ) = cos(β-γ-α+δ) [same angle, different sign doesn't matter]</p><p>• cos(γ-α-β+δ) = cos(α-β-γ+δ)</p><p>All cosine terms in the first brackets cancel out when added.</p><p>The second bracket terms also exhibit symmetry leading to cancellation.</p><p><strong>Therefore: x + y + z = 0</strong> ✓</p><p><strong>Step 4: Verify the Cubic Identity</strong></p><p>When x + y + z = 0, we have the algebraic identity:</p><p>If a + b + c = 0, then a³ + b³ + c³ = 3abc</p><p>Proof: From x + y + z = 0, we get z = -(x+y)</p><p>x³ + y³ + z³ = x³ + y³ - (x+y)³</p><p>= x³ + y³ - (x³ + 3x²y + 3xy² + y³)</p><p>= -3x²y - 3xy²</p><p>= -3xy(x+y)</p><p>= -3xy(-z)</p><p>= 3xyz ✓</p><p><strong>∴ Answer: a,b</strong></p>
Correct Answer: a,b