SOLUTION: 3cos(7x+1) = 3cos(1)(cos(7x)) - 3sin(1)(sin7x)
3sin(5x+3) = 3sin(3)(cos(5x)) + 3cos(3)(sin5x)
Now reverse these formulas and given the expanded version find the version with ju
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-> SOLUTION: 3cos(7x+1) = 3cos(1)(cos(7x)) - 3sin(1)(sin7x)
3sin(5x+3) = 3sin(3)(cos(5x)) + 3cos(3)(sin5x)
Now reverse these formulas and given the expanded version find the version with ju
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Question 1183303: 3cos(7x+1) = 3cos(1)(cos(7x)) - 3sin(1)(sin7x)
3sin(5x+3) = 3sin(3)(cos(5x)) + 3cos(3)(sin5x)
Now reverse these formulas and given the expanded version find the version with just one term. This involves solving a pair of equations -- in order to get Acos(x) +Bsin(x) = Rsin(x+b) = Rsin(b)cos(x)+Rcos(b)sin(x) what values must you choose for R and b? (Match coefficients.)
By convention we'll assume that the amplitude (the first coefficient on the left hand side) is positive.
_____ cos(7x+_____) = 7cos(7x)+1sin(7x)
_____sin(5x+______)=6cos(5x)=-3sin(5x) Answer by robertb(5830) (Show Source):
