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Factorised using the factor formulae, $\cos\theta-\cos3\theta-\cos5\theta+\cos7\theta$ equals:
A$4\cos4\theta\sin2\theta\sin\theta$
B$-4\sin4\theta\cos2\theta\cos\theta$
C$-4\cos4\theta\sin2\theta\sin\theta$
D$4\cos4\theta\cos2\theta\sin\theta$
Answer & Solution
Correct answer: C. $-4\cos4\theta\sin2\theta\sin\theta$
1. Group the terms so each pair is a sum: $(\cos7\theta+\cos\theta)-(\cos5\theta+\cos3\theta)$.
2. Apply $\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$ to both pairs.
3. This gives $2\cos4\theta\cos3\theta-2\cos4\theta\cos\theta$.
4. Factor out $2\cos4\theta$: $2\cos4\theta(\cos3\theta-\cos\theta)$.
5. Now $\cos3\theta-\cos\theta=-2\sin2\theta\sin\theta$, so the product is $-4\cos4\theta\sin2\theta\sin\theta$.
_Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 5: Trigonometry_
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