Precalculus Question of the Day

1. $\dfrac{\sqrt{6}-\sqrt{2}}{4}$
2. $\dfrac{\sqrt{6}+\sqrt{2}}{4}$ (correct answer)
3. $\dfrac{\sqrt{3}+1}{2}$
4. $\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}$

Explanation: This question tests understanding of the angle addition and subtraction formulas for sine. The sine addition formula is sin(A + B) = sin(A)cos(B) + cos(A)sin(B), which shows that sine of a sum is NOT simply sin(A) + sin(B) but requires cross terms involving both sine and cosine of each angle. To find sin(75°), we recognize 75° = 45° + 30°, so sin(75°) = sin(45°)cos(30°) + cos(45°)sin(30°) = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6 + √2)/4. Choice B is correct because it properly substitutes the angle values and simplifies accurately. Choice A has the wrong sign between the terms, using minus when the formula for sine addition requires plus. To use these formulas for exact values, break unfamiliar angles into sums or differences of standard angles (like 75° = 45° + 30° or 15° = 45° - 30°), then apply the formula with the known exact trig values. Common error: students try sin(A + B) = sin(A) + sin(B), but you can quickly verify this is wrong by trying A = B = 45°: sin(90°) = 1 but sin(45°) + sin(45°) = √2/2 + √2/2 = √2 ≠ 1.