Question 70660

Concept: For u ∈ [−1, 1] the value cos−1u is the angle in [0, π] whose cosine is u, so its sine is √(1 − u2) and is never negative. Two such angles are tied together by the cosine addition identity cos(A + B) = cos A cos B − sin A sin B. Squaring that identity is the standard way to eliminate the two sine radicals and leave a relation purely in x, y and cos φ.

Application — apply the identity to this expression:

1. Set A = cos−1(x/2) and B = cos−1(y/3), so cos A = x/2, cos B = y/3 and A + B = φ.

2. Expand cos(A + B): cos φ = (x/2)(y/3) − sin A sin B, which rearranges to sin A sin B = xy/6 − cos φ.

3. Square both sides to remove the radicals: sin2A · sin2B = (xy/6 − cos φ)2.

4. Substitute sin2A = 1 − x2/4 and sin2B = 1 − y2/9: (1 − x2/4)(1 − y2/9) = x2y2/36 − (xy/3)cos φ + cos2φ.

5. Expand the left side to 1 − x2/4 − y2/9 + x2y2/36 and cancel the common x2y2/36, leaving 1 − x2/4 − y2/9 = cos2φ − (xy/3)cos φ.

6. Multiply every term by 36 to clear the denominators: 36 − 9x2 − 4y2 = 36cos2φ − 12xy cos φ.

7. Rearrange so the required combination stands alone: 9x2 − 12xy cos φ + 4y2 = 36 − 36cos2φ = 36(1 − cos2φ) = 36 sin2φ.

Cross-check — test the result on two admissible pairs:

• x = 2, y = 0 gives A = 0, B = π/2, so φ = π/2 and cos φ = 0. The expression equals 9(4) − 0 + 0 = 36, and 36 sin2(π/2) = 36 as well.

• x = 2, y = 3 gives A = B = 0, so φ = 0 and cos φ = 1. The expression equals 36 − 72 + 36 = 0, and 36 sin20 = 0 as well; a constant value of 36 would fail this pair, and 36 cos2φ would give 36 instead of 0.

Therefore 9x2 − 12xy cos φ + 4y2 = 36 sin2φ.