Trigonometry & Inverse Trigonometry
Trigonometric Identities
Grade 11
Question:
<p>For statement \(p\): \(\theta = 240^\circ\), consider<br>\[2\sin\left(\frac{240^\circ}{2}\right) = \sqrt{1+\sin 240^\circ} - \sqrt{1-\sin 240^\circ}\]<br>Is statement \(p\) true or false, and what about statement \(q\): \(\cos\left(\frac{1}{2}(A+C)\right) + \cos\left(\frac{1}{2}(B+D)\right) = 0\)?</p>
<p>(1) Statement \(p\) is true and statement \(q\) is false</p>
<p>(2) Both statements \(p\) and \(q\) are true</p>
<p>(3) Both statements \(p\) and \(q\) are false</p>
<p>(4) Statement \(p\) is false and statement \(q\) is true</p>
Step-by-Step Solution
Key Concept: Recognize that √(1±sin θ) can be rewritten as |sin(θ/2) ± cos(θ/2)| using the identity (sin(θ/2) ± cos(θ/2))² = 1 ± sin θ, then evaluate carefully with angle quadrants.
<p><strong>Step 1:</strong> For statement p with θ = 240°, use the identity (sin(θ/2) ± cos(θ/2))² = 1 ± sin θ.</p><p>√(1 + sin 240°) = |sin 120° + cos 120°| = |√3/2 - 1/2| = (√3 - 1)/2</p><p>√(1 - sin 240°) = |sin 120° - cos 120°| = |√3/2 + 1/2| = (√3 + 1)/2</p><p><strong>Step 2:</strong> RHS = (√3 - 1)/2 - (√3 + 1)/2 = -1</p><p>LHS = 2sin(120°) = 2 · √3/2 = √3</p><p>Since √3 ≠ -1, statement p is <strong>FALSE</strong>.</p><p><strong>Step 3:</strong> For statement q about cos((A+C)/2) + cos((B+D)/2) = 0: Without additional constraints, this is generally <strong>FALSE</strong> for arbitrary angles A, B, C, D. It would only hold if A+C = π and B+D = π (as in a cyclic quadrilateral context), which isn't universally given.</p><p>∴ Answer: D</p>
Correct Answer: D