Use identities to find the exact values at A for the remaining five trigonometric functions
Cos(A)= -(sqr of 3, then divided by 6) and A is in the third quadrant
in QIII, x<0 and y<0
so, we have
x = -√3
r = 6
so y = √33
Now you just remember that
cosA = x/r
sinA = y/r
tanA = y/x
etc.
rats -
y = -√33
To find the values of trigonometric functions at angle A in the third quadrant when given the cosine value, we can use the Pythagorean identity and the trigonometric ratios in the unit circle.
Given that cos(A) = -√3 / 6, we can start by finding the value of sin(A) using the Pythagorean identity: sin^2(A) + cos^2(A) = 1.
cos(A) = -√3 / 6
cos^2(A) = (√3 / 6)^2 = 3 / 36 = 1 / 12
Using the Pythagorean identity, sin^2(A) = 1 - cos^2(A)
sin^2(A) = 1 - 1 / 12
sin^2(A) = 11 / 12
Since angle A is in the third quadrant, sin(A) is negative. Taking the square root and including the negative sign, we find:
sin(A) = -√(11/12) = -√11 / √12 = -√(11) / 2√3
Now, we have the values of cos(A) and sin(A). To find the remaining trigonometric functions, we can use the definitions:
tan(A) = sin(A) / cos(A)
cot(A) = 1 / tan(A)
sec(A) = 1 / cos(A)
csc(A) = 1 / sin(A)
Let's calculate these:
tan(A) = (sin(A) / cos(A)) = (-√(11) / 2√3) / (-√3/6)
= -√(11) / 2√3 * -6/√3
= -√11 / 2 * -6
= 6√11 / 6
= √11
cot(A) = 1 / tan(A) = 1 / √11
sec(A) = 1 / cos(A) = 1 / (-√3 / 6)
= -6 / √3
= -2√3
csc(A) = 1 / sin(A) = 1 / (-√(11) / 2√3)
= -2√3 / √11
= -2√3√11 / 11
= -2√(33) / 11
Therefore, the values of the remaining five trigonometric functions at angle A in the third quadrant are:
tan(A) = √11
cot(A) = 1 / √11
sec(A) = -2√3
csc(A) = -2√(33) / 11