Trigonometry & Inverse Trigonometry Question - None

Question

If x\ and y\ are acute angles, such that \cos x + \cos y = \frac{3}{2}\ and \sin x + \sin y = \frac{3}{4}\, then \sin(x + y)\ equals

Accepted Answer

Step 1: Square the first equation: (cos x + cos y)² = (3/2)²

cos²x + cos²y + 2cos x cos y = 9/4 ... (i)

Step 2: Square the second equation: (sin x + sin y)² = (3/4)²

sin²x + sin²y + 2sin x sin y = 9/16 ... (ii)

Step 3: Add equations (i) and (ii):

(cos²x + sin²x) + (cos²y + sin²y) + 2(cos x cos y + sin x sin y) = 9/4 + 9/16

1 + 1 + 2cos(x - y) = 36/16 + 9/16 = 45/16

2 + 2cos(x - y) = 45/16

2cos(x - y) = 45/16 - 2 = 45/16 - 32/16 = 13/16

cos(x - y) = 13/32 ... (iii)

Step 4: Use sum-to-product formulas. From cos x + cos y = 3/2:

2cos((x+y)/2)cos((x-y)/2) = 3/2 ... (iv)

From sin x + sin y = 3/4:

2sin((x+y)/2)cos((x-y)/2) = 3/4 ... (v)

Step 5: Divide equation (v) by equation (iv):

tan((x+y)/2) = (3/4)/(3/2) = (3/4) × (2/3) = 1/2

Step 6: Using tan((x+y)/2) = 1/2, apply the double angle formula:

sin(x+y) = 2sin((x+y)/2)cos((x+y)/2) = 2tan((x+y)/2)/(1 + tan²((x+y)/2))

sin(x+y) = 2(1/2)/(1 + 1/4) = 1/(5/4) = 4/5

Step 7: Verify using cos(x+y). Since tan((x+y)/2) = 1/2, we have cos((x+y)/2) = 2/√5 and sin((x+y)/2) = 1/√5.

cos(x+y) = cos²((x+y)/2) - sin²((x+y)/2) = 4/5 - 1/5 = 3/5

Check: sin²(x+y) + cos²(x+y) = (4/5)² + (3/5)² = 16/25 + 9/25 = 25/25 = 1 ✓

∴ Answer: C

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