Trigonometry & Inverse Trigonometry
Trigonometric Identities and Series
Grade 11
Question:
<p>If \(\cos^3 x \sin 2x = \sum_{r=0}^{n} a_r \sin(rx)\), \(\forall x \in R\) then</p>
<p>(a) \(n = 5, a_1 = \dfrac{1}{2}\)</p>
<p>(b) \(n = 5, a_1 = \dfrac{1}{4}\)</p>
<p>(c) \(n = 5, a_2 = \dfrac{1}{8}\)</p>
<p>(d) \(n = 5, a_2 = \dfrac{1}{4}\)</p>
Step-by-Step Solution
Key Concept: Express cos³x·sin(2x) using product-to-sum formulas and power reduction identities, then decompose into a linear combination of sin(rx) terms to identify the coefficient pattern and find n.
<p><strong>Step 1:</strong> Rewrite using sin(2x) = 2sin(x)cos(x):<br/>cos³x·sin(2x) = cos³x·2sin(x)cos(x) = 2sin(x)cos⁴x</p><p><strong>Step 2:</strong> Use power reduction on cos⁴x:<br/>cos⁴x = ((1+cos(2x))/2)² = (1 + 2cos(2x) + cos²(2x))/4<br/>= (1 + 2cos(2x) + (1+cos(4x))/2)/4 = (3 + 4cos(2x) + cos(4x))/8</p><p><strong>Step 3:</strong> Multiply by 2sin(x):<br/>2sin(x)·(3 + 4cos(2x) + cos(4x))/8 = (1/4)[3sin(x) + 4sin(x)cos(2x) + sin(x)cos(4x)]</p><p><strong>Step 4:</strong> Apply product-to-sum: sin(A)cos(B) = [sin(A+B) + sin(A-B)]/2<br/>= (1/4)[3sin(x) + 2(sin(3x) + sin(-x)) + (1/2)(sin(5x) + sin(-3x))]<br/>= (1/4)[3sin(x) + 2sin(3x) - 2sin(x) + (1/2)sin(5x) - (1/2)sin(3x)]<br/>= (1/4)[sin(x) + (3/2)sin(3x) + (1/2)sin(5x)]</p><p><strong>Step 5:</strong> Identify that highest power is sin(5x), so n = 5 and coefficients are a₀=0, a₁=1/4, a₃=3/8, a₅=1/8, others=0</p><p>∴ Answer: B</p>
Correct Answer: B