evaluate sec (7π/6)
7(180) /6 = 210
the answer is - √ 3 / 2
if it's not correct, can you correct me with an explanation.. :/ i'm so lost, thanks!
7π/6 is (I assume) an angle in radians. It is equivalent to 210 degrees.
sec (secant) is a trigonometric function, equivalent to 1 / cos
An angle of 7π/6 (or 210 degrees), falls in the third quadrant (30 degrees past 180).
A unit vector pointing in that direction will have an x value of -√3/2 and a y value of -(1/2).
The "unit vector" has a length of 1 and forms the "hypotenuse" From the centre of the graph, the adjacent side is along the x-axis.
cosine = adjacent(x) / hypotenuse = (-√3 / 2)/1 = -√3 / 2 = -0.866...
secant = 1/cos = - 2/√3 = -1.1547
sec (7Ï/6) = sec (Ï + Ï/6) = -sec(Ï/6) = -2/sqrt(3) = -(2/3)sqrt(3)
sec (7Ï/6) = sec {Ï+(Ï/6)} = - sec {(Ï/6)} = - {1/ cos(Ï/6)} =-{1/( â 3 / 2)} = -{2/ â 3}.......ANS
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7π/6 is (I assume) an angle in radians. It is equivalent to 210 degrees.
sec (secant) is a trigonometric function, equivalent to 1 / cos
An angle of 7π/6 (or 210 degrees), falls in the third quadrant (30 degrees past 180).
A unit vector pointing in that direction will have an x value of -√3/2 and a y value of -(1/2).
The "unit vector" has a length of 1 and forms the "hypotenuse" From the centre of the graph, the adjacent side is along the x-axis.
cosine = adjacent(x) / hypotenuse = (-√3 / 2)/1 = -√3 / 2 = -0.866...
secant = 1/cos = - 2/√3 = -1.1547
sec (7Ï/6) = sec (Ï + Ï/6) = -sec(Ï/6) = -2/sqrt(3) = -(2/3)sqrt(3)
sec (7Ï/6) = sec {Ï+(Ï/6)} = - sec {(Ï/6)} = - {1/ cos(Ï/6)} =-{1/( â 3 / 2)} = -{2/ â 3}.......ANS