SOLUTION: Which of the following trigonometric functions of x is not correct given that sin x > 0 and sec x = -2? • csc x = ( 2 sq rt 3) / 3 • cos x = - 1 / 2 • cot x = - ( sq rt 3 ) / 3

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Which of the following trigonometric functions of x is not correct given that sin x > 0 and sec x = -2? • csc x = ( 2 sq rt 3) / 3 • cos x = - 1 / 2 • cot x = - ( sq rt 3 ) / 3       Log On


   

Question 3656: Which of the following trigonometric functions of x is not correct given that sin x > 0 and sec x = -2?
• csc x = ( 2 sq rt 3) / 3
• cos x = - 1 / 2
• cot x = - ( sq rt 3 ) / 3
• tan x = - ( sq rt 3 ) / 2
Answer by Zenock(15) About Me  (Show Source):
You can put this solution on YOUR website!
I'm not sure why this is under "Quadratic Equations" but oh well here we go.
First let's define the following in terms of sin(x) and cos(x):
+csc%28x%29=1%2Fsin%28x%29 so sin%28x%29=1%2Fcsc%28x%29
+sec%28x%29=1%2Fcos%28x%29+ so cos%28x%29=1%2Fsec%28x%29
+tan%28x%29=sin%28x%29%2Fcos%28x%29+
+cot%28x%29=1%2Ftan%28x%29=cos%28x%29%2Fsin%28x%29
Additionally we need to note that:
+sin%5E2++%28x%29+%2B+cos%5E2%28x%29=1+
Now we begin.
First lets find the cos(x):
cos%28x%29=1%2Fsec%28x%29=1%2F-2=-+1%2F2 so cos x = - 1 / 2 is correct
Now lets find the sin(x) (Not in our list but needed):
sin%5E2%28x%29%2Bcos%5E2%28x%29=1 so sin%5E2%28x%29%2B+1%2F4=1 so sin%5E2%28x%29=3%2F4
Therefore: +sin%28x%29+=+-sqrt%283%29%2F2+ or +sin%28x%29+=+sqrt%283%29%2F2+
However from the orginal problem: sin%28x%29%3E0 so +sin%28x%29+=+sqrt%283%29%2F2+
Using this information:
csc%28x%29=1%2Fsin%28x%29=1%2F%28sqrt%283%29%2F2%29=2%2Fsqrt%283%29=2sqrt%283%29%2F3 so csc x = ( 2 sq rt 3) / 3 is correct
so cot x = - ( sq rt 3 ) / 3 is correct
tan%28x%29=sin%28x%29%2Fcos%28x%29=%28sqrt%283%29%2F2%29%2F%28-+1%2F2%29=-sqrt%283%29 so tan x = - ( sq rt 3 ) / 2 is **** not correct ****