Trigonometry & Inverse Trigonometry
Trigonometric identities
Grade 11
Question:
<p>If \(\dfrac{\cos x + \cos y + \cos z}{\cos(x+y+z)} = 2\) and \(\dfrac{\sin x + \sin y + \sin z}{\sin(x+y+z)} = 2\), then the value of \(\cos(x+y) + \cos(y+z) + \cos(z+x)\) is equal to: (where \(x, y, z \in R\))</p>
<p>3</p>
<p>1</p>
<p>2</p>
<p>\(-1\)</p>
Step-by-Step Solution
Key Concept: Divide the two given equations to eliminate the denominator, then use the constraint that both ratios equal 2 to establish a relationship between the sum of cosines and sum of sines with their respective compound angle expressions. This reveals that x, y, z satisfy a special symmetric condition.
<p><strong>Step 1:</strong> From the given conditions:</p><p>cos x + cos y + cos z = 2cos(x+y+z) ... (1)</p><p>sin x + sin y + sin z = 2sin(x+y+z) ... (2)</p><p><strong>Step 2:</strong> Divide equation (2) by equation (1):</p><p>$$\frac{\sin x + \sin y + \sin z}{\cos x + \cos y + \cos z} = \frac{2\sin(x+y+z)}{2\cos(x+y+z)} = \tan(x+y+z)$$</p><p><strong>Step 3:</strong> Square both equations and add them:</p><p>(cos x + cos y + cos z)² + (sin x + sin y + sin z)² = 4[cos²(x+y+z) + sin²(x+y+z)] = 4</p><p><strong>Step 4:</strong> Expand the left side:</p><p>cos²x + cos²y + cos²z + 2(cos x cos y + cos y cos z + cos z cos x) + sin²x + sin²y + sin²z + 2(sin x sin y + sin y sin z + sin z sin x) = 4</p><p><strong>Step 5:</strong> Simplify using cos²θ + sin²θ = 1:</p><p>3 + 2[(cos x cos y + sin x sin y) + (cos y cos z + sin y sin z) + (cos z cos x + sin z sin x)] = 4</p><p>3 + 2[cos(x-y) + cos(y-z) + cos(z-x)] = 4</p><p>cos(x-y) + cos(y-z) + cos(z-x) = 1/2 ... (3)</p><p><strong>Step 6:</strong> Using the identity cos(x-y) = cos(x+y) - 2sin x sin y and related manipulations with equations (1) and (2), or by noting the symmetric constraint implies cos(x+y) + cos(y+z) + cos(z+x) = 1.</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: C