Describe the possible configurations and the possible number of points of intersection of (a) 2 lines in the plane (b) 3 lines i
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Givens
y = 2
x = 1
z(the hypotenuse) = √(2^2 + 1^2) = √5
Cos(u) = x value / hypotenuse = 1/√5
Sin(u) = y value / hypotenuse = 2/√5
Solve for sin2u
Sin(2u) = 2*sin(u)*cos(u)
Sin(2u) = 2() = 4/5
Solve for cos(2u)
cos(2u) = - sqrt(1 - sin^2(2u))
Cos(2u) = - sqrt(1 - (4/5)^2 )
Cos(2u) = -sqrt(1 - 16/25)
cos(2u) = -sqrt(9/25)
cos(2u) = -3/5
Solve for Tan(2u)
tan(2u) = sin(2u) / cos(2u) = 4/5// - 3/5 = - 0.8/0.6 = - 1.3333 = - 4/3
Notes
One: Notice that you would normally rationalize the denominator, but you don't have to in this case. The formulas are such that they perform the rationalizations themselves.
Two: Notice the sign on the cos(2u). The sin is plus even though the angle (2u) is in the second quadrant. The cos is different. It is about 126 degrees which would make it a negative root (9/25)
Three: If you are uncomfortable with the tan, you could do fractions.