To evaluate sin(165°), you write it as sin(330°/2). Before applying the formula sin(θ/2) = ±√((1-cosθ)/2), what must you determine first?
AThe quadrant of 330°, to find the correct value of cos(330°)
BThe quadrant of 165°, to determine the correct sign of sin(165°)
CWhether to use the sine or cosine half-angle formula
DWhether 330° has a standard exact cosine value
The ± sign is determined by the quadrant of θ/2, not θ. Here θ/2 = 165°, which lies in the second quadrant, where sine is positive — so the + sign applies. Knowing the quadrant of θ = 330° helps compute cos(330°), but the sign decision depends specifically on the quadrant of the half-angle 165°.
Question 2 Multiple Choice
A student claims that since sin(60°) = √3/2, it follows that sin(30°) = (√3/2)/2 = √3/4. Which statement best explains the error?
AThe arithmetic is wrong; √3/4 is not equal to (√3/2)/2
BThe formula sin(θ/2) = sin(θ)/2 only works when θ/2 is in the first quadrant
CSine is not a linear function; the correct formula sin(θ/2) = ±√((1-cosθ)/2) gives sin(30°) = √((1-cos60°)/2) = √(1/4) = 1/2
DHalf-angle formulas require using the cosine of the full angle, so the student should have used cos(60°)
The misconception is treating sine as linear — halving the input does not halve the output. The correct formula, derived from cos(θ) = 1 - 2sin²(θ/2), gives sin(θ/2) = ±√((1-cosθ)/2). For θ = 60°: sin(30°) = √((1 - 1/2)/2) = √(1/4) = 1/2, not √3/4. Option D correctly notes that cos(60°) is needed, but misses that the whole approach of dividing by 2 is wrong.
Question 3 True / False
The ± sign in sin(θ/2) = ±√((1-cosθ)/2) is determined by the quadrant of θ, not the quadrant of θ/2.
TTrue
FFalse
Answer: False
The sign is determined by the quadrant of θ/2, the half-angle itself. The formula gives the value of sin(θ/2), so its sign depends on which quadrant θ/2 occupies. Example: sin(75°) = sin(150°/2). θ = 150° is in Q2, but what matters is θ/2 = 75° is in Q1, where sine is positive.
Question 4 True / False
For any angle θ in the first quadrant (0° < θ < 90°), both sin(θ/2) and cos(θ/2) are positive.
TTrue
FFalse
Answer: True
If 0° < θ < 90°, then 0° < θ/2 < 45°. An angle between 0° and 45° is in Q1, where both sine and cosine are positive. So the + sign applies to both half-angle formulas without ambiguity. The sign complication arises only when θ/2 falls in Q2, Q3, or Q4.
Question 5 Short Answer
Explain why sin(15°) cannot be found by computing sin(30°)/2, and describe the correct approach.
Think about your answer, then reveal below.
Model answer: Sine is not linear, so sin(15°) ≠ sin(30°)/2. The correct approach uses the half-angle identity: sin(15°) = sin(30°/2) = ±√((1-cos30°)/2). Since 15° is in Q1, the + sign applies: sin(15°) = √((1 - √3/2)/2) = √((2-√3)/4) = √(2-√3)/2 ≈ 0.259.
For a linear function, halving the input halves the output. But sin(30°)/2 = (1/2)/2 = 0.25, while the actual sin(15°) ≈ 0.259 — close but wrong. The half-angle formula is derived algebraically from the double-angle identity and correctly accounts for the nonlinear shape of the sine curve. The error compounds for other angles.