SOLUTION: I need help with solving the following:
tan2x = 2tan / 1- tan^2 , x = pi/6
tan2(pi/6) = 2 tan (pi/6) / 1- tan(pi/6)^2
tan square root 3 = 2 tan (pi/6) / 1 - tan (pi/6)^2
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Question 737432: I need help with solving the following:
tan2x = 2tan / 1- tan^2 , x = pi/6
tan2(pi/6) = 2 tan (pi/6) / 1- tan(pi/6)^2
tan square root 3 = 2 tan (pi/6) / 1 - tan (pi/6)^2
tan square root 3 = 2 tan square root 3 / 1- (square root 3/2)
and this is where I get lost at
Let me know if I did it correctly
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
at this point, is pi/6 degrees or radians? I assume radians
tan2(pi/6) = 2 tan (pi/6) / 1- tan(pi/6)^2
evaluate the right side of the = sign
2*tan(3.14/6) / (1-tan(3.14/6)^2
2*.58 / (1-.58^2) = 1.16 / .66 = .5
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