<p>Represent <span>\(\cos 6\theta\)</span> in terms of <span>\(\cos \theta\)</span>.</p>
Step-by-Step Solution
Key Concept: Use De Moivre's theorem: (cos θ + i sin θ)⁶ = cos 6θ + i sin 6θ, then expand the left side using the binomial theorem and equate real parts. The real part of the expansion gives cos 6θ as a polynomial in cos θ and sin θ, which can be converted entirely to cos θ using sin²θ = 1 − cos²θ.
<p><strong>Step 1:</strong> Apply De Moivre's Theorem</p><p>(cos θ + i sin θ)⁶ = cos 6θ + i sin 6θ</p><p><strong>Step 2:</strong> Expand left side using Binomial Theorem</p><p>(cos θ + i sin θ)⁶ = Σ C(6,r) cos⁶⁻ʳθ (i sin θ)ʳ</p><p><strong>Step 3:</strong> Extract real part (r = 0, 2, 4, 6 terms)</p><p>cos 6θ = C(6,0)cos⁶θ − C(6,2)cos⁴θ sin²θ + C(6,4)cos²θ sin⁴θ − C(6,6)sin⁶θ</p><p>cos 6θ = cos⁶θ − 15cos⁴θ sin²θ + 15cos²θ sin⁴θ − sin⁶θ</p><p><strong>Step 4:</strong> Substitute sin²θ = 1 − cos²θ</p><p>cos 6θ = cos⁶θ − 15cos⁴θ(1 − cos²θ) + 15cos²θ(1 − cos²θ)² − (1 − cos²θ)³</p><p><strong>Step 5:</strong> Expand and simplify</p><p>= cos⁶θ − 15cos⁴θ + 15cos⁶θ + 15cos²θ(1 − 2cos²θ + cos⁴θ) − (1 − 3cos²θ + 3cos⁴θ − cos⁶θ)</p><p>= cos⁶θ − 15cos⁴θ + 15cos⁶θ + 15cos²θ − 30cos⁴θ + 15cos⁶θ − 1 + 3cos²θ − 3cos⁴θ + cos⁶θ</p><p>= 32cos⁶θ − 48cos⁴θ + 18cos²θ − 1</p><p><strong>∴ Answer: cos 6θ = 32cos⁶θ − 48cos⁴θ + 18cos²θ − 1</strong></p>
Correct Answer: 32cos⁶θ − 48cos⁴θ + 18cos²θ − 1