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AlladinOne [14]
3 years ago
9

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

Mathematics
1 answer:
timofeeve [1]3 years ago
6 0

Answer:

sin(2A) = (2√2 + √3) / 6

Step-by-step 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

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Answer:

The area of the triangle is 28 square units

Step-by-step explanation:

First we have to calculate the length of the base of the triangle and the length of the height

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