Trigonometry & Inverse Trigonometry
General Solutions of Trigonometric Equations
Grade 11
Question:
<p>If \(\cos A = \cos B\) and \(\sin A = \sin B\) then</p>
<p>(a) \(A + B = 0\)</p>
<p>(b) \(A - B = 0\)</p>
<p>(c) \(A + B = 2n\pi\)</p>
<p>(d) \(A = B + 2n\pi\)</p>
Step-by-Step Solution
Key Concept: Two angles have identical cosine and sine values if and only if they differ by a multiple of 2π. This is because the trigonometric functions are periodic with period 2π, and the pair (cos θ, sin θ) uniquely determines the angle modulo 2π.
<p><strong>Step 1:</strong> Recall that cos A = cos B has general solution A = ±B + 2πk (k ∈ ℤ).</p><p><strong>Step 2:</strong> Recall that sin A = sin B has general solution A = B + 2πk or A = π - B + 2πk (k ∈ ℤ).</p><p><strong>Step 3:</strong> Both conditions must be satisfied simultaneously. The only solution that satisfies BOTH is A = B + 2πk.</p><p><strong>Step 4:</strong> This can be verified: if A = B + 2πk, then cos A = cos(B + 2πk) = cos B ✓ and sin A = sin(B + 2πk) = sin B ✓</p><p><strong>Step 5:</strong> If A = π - B + 2πk, then sin A = sin B ✓ but cos A = cos(π - B) = -cos B ✗</p><p>∴ Answer: A = B + 2πk, where k ∈ ℤ</p>
Correct Answer: D