If cot x = ( - square root 2) and sin x > 0, find cos x
cotØ = -√2 and sinØ > 0
Ø must be in quad II by the CAST rule
cotØ = -√2 = x/y , so x = -√2 and y = 1 (in quad II)
x^2 + y^2 = r^2
2 + 1 = r^2
r = √3
cosØ = x/r = -√2/√3 or if you rationalize it, -√6/3
cotx = (cosx)/ (sinx). Since sin>0, it must be in quadrant I or II. Since cotx is negative, cosx<0 or negative. The only quadrant where sin>0 and cosx<0 is quadrant II. Using exact values of trigonometric functions, you can know that cosx = -(sqrt(2))/ 2 and sinx = 2/1.
as above, we know that θ is in QII. Unfortunately, sinθ is never 2/1. and cosθ is also wrong.
cotθ = x/y = -√2/1
r = √3
cosθ = x/r = -√2/√3
To find the value of cos(x), we can use the equation cos^2(x) + sin^2(x) = 1. We already know that sin(x) is greater than 0, but we need to find the value of sin(x).
Given that cot(x) = -√2, we can use the relationship between cot(x) and sin(x) to find the value of sin(x).
Cot(x) is the reciprocal of tan(x), so we can write cot(x) = 1/tan(x).
Since cot(x) = -√2, we can rewrite it as 1/tan(x) = -√2.
To find the value of tan(x), we can use the relationship between cot(x) and tan(x):
cot(x) = cos(x)/sin(x) = -√2.
By rearranging this equation, we can get cos(x) = -√2 * sin(x).
Now we can substitute cos(x) = -√2 * sin(x) into the equation cos^2(x) + sin^2(x) = 1:
(-√2 * sin(x))^2 + sin^2(x) = 1.
Simplifying this equation gives us:
2 * sin^2(x) + sin^2(x) = 1,
3 * sin^2(x) = 1,
sin^2(x) = 1/3.
Since sin(x) > 0, sin(x) = √(1/3) = 1/√3.
Now we can substitute sin(x) = 1/√3 into cos(x) = -√2 * sin(x):
cos(x) = -√2 * (1/√3) = -√(2/3).
Therefore, cos(x) = -√(2/3).