Complex Numbers
Cube Roots and Powers
Grade 11
Question:
<p>If <span class="math">\cos \alpha + \cos \beta + \cos \gamma = \sin \alpha + \sin \beta + \sin \gamma = 0</span>, then <span class="math">\cos 3\alpha + \cos 3\beta + \cos 3\gamma</span> is equal to</p>
<p>(a) 0</p>
<p>(b) <span class="math">\cos(\alpha + \beta + \gamma)</span></p>
<p>(c) <span class="math">3\cos(\alpha + \beta + \gamma)</span></p>
<p>(d) <span class="math">3\sin(\alpha + \beta + \gamma)</span></p>
Step-by-Step Solution
Key Concept: Represent the given conditions using complex numbers where cos θ + i sin θ = e^(iθ), then use the constraint that both real and imaginary parts sum to zero to find a relationship between the angles.
<p><strong>Step 1:</strong> Convert the given conditions to complex form. Let z₁ = e^(iα), z₂ = e^(iβ), z₃ = e^(iγ).</p><p>From the given conditions:<br/>cos α + cos β + cos γ = 0 (real parts)<br/>sin α + sin β + sin γ = 0 (imaginary parts)<br/>This means: <strong>e^(iα) + e^(iβ) + e^(iγ) = 0</strong></p><p><strong>Step 2:</strong> From e^(iα) + e^(iβ) + e^(iγ) = 0, we have:<br/>e^(iγ) = -(e^(iα) + e^(iβ))</p><p><strong>Step 3:</strong> Cube both sides:<br/>e^(i3γ) = -[e^(iα) + e^(iβ)]³</p><p><strong>Step 4:</strong> Expand the right side:<br/>e^(i3γ) = -(e^(i3α) + 3e^(i2α)e^(iβ) + 3e^(iα)e^(i2β) + e^(i3β))<br/>e^(i3γ) = -e^(i3α) - 3e^(i(2α+β)) - 3e^(i(α+2β)) - e^(i3β)</p><p><strong>Step 5:</strong> Add e^(i3α) + e^(i3β) + e^(i3γ):<br/>e^(i3α) + e^(i3β) + e^(i3γ) = e^(i3α) + e^(i3β) - e^(i3α) - 3e^(i(2α+β)) - 3e^(i(α+2β)) - e^(i3β)<br/>= -3[e^(i(2α+β)) + e^(i(α+2β))]<br/>= -3e^(i(α+β))[e^(iα) + e^(iβ)]<br/>= 3e^(i(α+β))e^(iγ)<br/>= 3e^(i(α+β+γ))</p><p><strong>Step 6:</strong> Taking the real part:<br/>cos 3α + cos 3β + cos 3γ = Re[3e^(i(α+β+γ))]<br/>= <strong>3cos(α + β + γ)</strong></p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C