sin^2 + cos^2 = 1 ... cos = √(1 - sin^2) = √[1 - (-2/5)^2]
find the cosine ... remember, it's positive in Quad IV
sec = 1 / cos ... tan = sin / cos
Let theta be an angle in quadrant IV such that sin(theta)=-(2)/(5).
Find the exact values of sec(theta) and tan(theta).
3 answers
in QIV, x is positive and y is negative. So, if sinθ = -2/5, you have
y = -2
r = 5
x = √21
Now, recall that
sinθ = y/r
cosθ = x/r
tanθ = y/x
and of course, there are the reciprocals: csc, sec, cot
y = -2
r = 5
x = √21
Now, recall that
sinθ = y/r
cosθ = x/r
tanθ = y/x
and of course, there are the reciprocals: csc, sec, cot
Y = -2.
r = 5.
x^2 + (-2)^2 = 5^2,
X = sqrt(21).
Sec A = 1/Cos A = 1 / (x/r) = r/x = 5/sqrt(21).
Tan A = Y/X = -2/sqrt(21).
r = 5.
x^2 + (-2)^2 = 5^2,
X = sqrt(21).
Sec A = 1/Cos A = 1 / (x/r) = r/x = 5/sqrt(21).
Tan A = Y/X = -2/sqrt(21).
30 answers