If sin 3y = cos 2y and 0o $\leq$ 90o, find the value of y

Analyze the question: We need y (between 0° and 90°) such that sin 3y = cos 2y.

Concept: The complementary angle identity says sin θ = cos(90° − θ). So if sin 3y = cos 2y, the angles 3y and 2y must be complementary (add up to 90°).

Solution:

sin 3y = cos 2y cos(90° − 3y) = cos 2y

So 90° − 3y = 2y 90° = 5y y = 18°

Check:

• sin 3(18°) = sin 54° ≈ 0.809
• cos 2(18°) = cos 36° ≈ 0.809 ✓

Other options: Is there another solution in the range? Using cos(90° − 3y) = cos 2y also gives 90° − 3y = −2y, so y = 90°. Check: sin 270° = −1, but cos 180° = −1. That also works, so y = 90° technically satisfies the equation if the range includes 90°. The question as printed is garbled ("0° ≤ 90°"), and the intended range is most likely 0° ≤ y < 90°, or the standard expected answer is the principal one from the complementary-angle method. Option B (36°): sin 108° ≈ 0.951 and cos 72° ≈ 0.309, so it fails. Option C (54°): sin 162° ≈ 0.309 and cos 108° ≈ −0.309, so it fails.

The expected answer is **A. Note that 90° also technically satisfies the equation if the range is inclusive of 90°, but the intended solution is the complementary-angle one.

Tip: When you see sin A = cos B, immediately think A + B = 90°. This saves time and works for most WASSCE/JAMB questions of this type.