a wheel has a radius of 1.4 feet and turns with an angular speed of 2.65 radians per second. Determine the distance traveled by
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Tan3A=tan(2A+A)
We know that , tan(x+y)=(tanx + tany)/(1 - (tanx)(tany))
tan(2A+A)=(tan2A+tanA)/(1 - (tan2A)(tanA))— (1)
We know that , tan2x=2tanx/(1 - tan^2x)
So, by substituting tan2A in (1),we get,
=[2tanA/(1 - tan^2A) + tanA]/1- (2tanA/(1 - tan^2A))(tanA)]
=[2tanA+tanA - tan^3A]/[1 - tan^2A - 2tan^2A]
=[3tanA - tan^3A]/[1-3tan^2A]
Therefore, tan3A= [3tanA - tan^3A]/[1-3tan^2A]
We know that , tan(x+y)=(tanx + tany)/(1 - (tanx)(tany))
tan(2A+A)=(tan2A+tanA)/(1 - (tan2A)(tanA))— (1)
We know that , tan2x=2tanx/(1 - tan^2x)
So, by substituting tan2A in (1),we get,
=[2tanA/(1 - tan^2A) + tanA]/1- (2tanA/(1 - tan^2A))(tanA)]
=[2tanA+tanA - tan^3A]/[1 - tan^2A - 2tan^2A]
=[3tanA - tan^3A]/[1-3tan^2A]
Therefore, tan3A= [3tanA - tan^3A]/[1-3tan^2A]