How Do You Isolate A in the Equation n=sin[(A+B)/2]/sin(B/2)?

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To isolate A in the equation n=sin[(A+B)/2]/sin(B/2), the user has transformed the equation to 2nsin(B/2) = sin(A+B). They recognize that this can be expanded to sinAcosB + cosAsinB, but find it unhelpful for their goal. A suggestion is made to consider the identity sin(B/2) = (1-cos(B))/2, which may provide further insights. The discussion indicates a need for additional strategies or identities to effectively isolate A. The conversation revolves around algebraic manipulation and trigonometric identities to solve the problem.
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Homework Statement



given this formula, i have to solve for A

n=sin[(A+B)/2]/sin(B/2)


Homework Equations





The Attempt at a Solution



Heres how far I got

2nsin(B/2) = sin(A+B) i know this can expand out to
=sinAcosB + cosAsinB but this isn't much help. Anyone got a suggestion
 
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joemama69 said:

Homework Statement



given this formula, i have to solve for A

n=sin[(A+B)/2]/sin(B/2)


Homework Equations





The Attempt at a Solution



Heres how far I got

2nsin(B/2) = sin(A+B) i know this can expand out to
=sinAcosB + cosAsinB but this isn't much help. Anyone got a suggestion

Made me think of the interweb

sin(\frac{B}{2}) = sin(\frac{B}{2}) = \frac{1-cos(B)}{2}
 
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