What is the final term in the distance formula for maximum range?

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The discussion centers on finding the angle alpha for maximum range in a projectile motion formula. The user is confused about the term cos(2alpha-beta) in the derivative of the distance formula. Another participant suggests that this term may be a misprint and explains its relation to the cosine addition formula. The clarification helps resolve the user's confusion about the derivation process. The conversation highlights the importance of accurate notation in mathematical expressions.
delve
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Hey guys,

I was hoping somebody might help me; I'm really confused. I obtained the formula for distance, which is this:

d=[(2[v_0]^2(cos[alpha])sin([alpha-beta])]/gcos^2(beta)

And the derivative for distance, which is this:

derivative of d with respect to alpha=0=(2[v_0]^2)/(gcos^2(beta))*(-sin(alpha)sin(alpha-beta)+cos(alpha)cos(alpha-beta))*cos(2alpha-beta)

I was trying to see what alpha would be for maximum range, but I have no idea where the very end term on the right comes from: cos(2alpha-beta). Could somebody please me? Thank you.

David
 
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Hey David! :smile:
delve said:
d=[(2[v_0]^2(cos[alpha])sin([alpha-beta])]/gcos^2(beta)

And the derivative for distance, which is this:

derivative of d with respect to alpha=0=(2[v_0]^2)/(gcos^2(beta))*(-sin(alpha)sin(alpha-beta)+cos(alpha)cos(alpha-beta))*cos(2alpha-beta)

No, that *cos(2α-β) at the end must be a misprint for =cos(2α-β) …

cos(A+B) = cosAcosB - sinAsinB :wink:
 
Thank you very much Tiny-Tim! That was exactly what I needed! :D
 
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