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Given sin(a) = 7/9 and a is in quadrant I, find the exact value of sin(a/2).

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User Ecarrion
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7 votes

Answer:

We can use the half-angle formula for sine to find the exact value of sin(a/2) in terms of sin(a):

sin(a/2) = ±√[(1 - cos(a))/2]

where the ± sign depends on the quadrant of a/2.

To use this formula, we first need to find cos(a). We can do this using the identity:

sin^2(a) + cos^2(a) = 1

Substituting sin(a) = 7/9, we get:

(7/9)^2 + cos^2(a) = 1

Simplifying and solving for cos(a), we get:

cos(a) = ±4/9

Since a is in quadrant I, we take the positive value of cos(a):

cos(a) = 4/9

Now we can use the half-angle formula for sine to find sin(a/2):

sin(a/2) = ±√[(1 - cos(a))/2]

Substituting cos(a) = 4/9, we get:

sin(a/2) = ±√[(1 - 4/9)/2]

Simplifying, we get:

sin(a/2) = ±√(5/18)

Since a is in quadrant I, a/2 is also in quadrant I, so we take the positive value of sin(a/2):

sin(a/2) = √(5/18)

Therefore, the exact value of sin(a/2) is √(5/18)

User ERIZ
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