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If sin(A+B)=1/2 & sin(A-B)=1/3 find sin(2A)

User Cogslave
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1 Answer

3 votes

Answer:

sin(2A) = (2√2 + √3) / 6

Explanation:

2A = (A+B) + (A−B)

sin(2A) = sin((A+B) + (A−B))

Angle sum formula:

sin(2A) = sin(A+B) cos(A−B) + sin(A−B) cos(A+B)

sin(2A) = 1/2 cos(A−B) + 1/3 cos(A+B)

Pythagorean identity:

sin(2A) = 1/2 √[1 − sin²(A−B)] + 1/3 √[1 − sin²(A+B)]

sin(2A) = 1/2 √(1 − 1/9) + 1/3 √(1 − 1/4)

sin(2A) = 1/2 √(8/9) + 1/3 √(3/4)

sin(2A) = 1/3 √2 + 1/6 √3

sin(2A) = (2√2 + √3) / 6

User Gabbar
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