Mathematics, 29.06.2019 11:10 olly09
Tan θ = y over x tan 11 pi over 6 = the fraction negative 1 over 2 over the fraction square root 3 over 2 tan 11 pi over 6 = negative 1 over 2 times 2 over square root 3 tan 11 pi over 6 = negative 1 over square root 3 tan 11 pi over 6 = negative square root 3 over 3 this is an example on my algebra 2 of the tangent of a point on the unit circle. can anyone tell me how negative 1 over the square root of 3 simplifies to negative square root of 3 over 3?
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Tan θ = y over x tan 11 pi over 6 = the fraction negative 1 over 2 over the fraction square root 3 o...
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