The question was:
"Given that sin theta is -1/2 and theta is in quadrant IV, find the cos and tan of theta."
I found 1/2 for cosine and -1 for theta, but the answer sheet said I was incorrect. I am almost positive I did this correctly... can someone double check for me?
make a triangle, and you will see that
y = -1, x = sqrt(3) and r = 2
so cos(theta) = sqrt(3)/2
tan(theta) = -1/sqrt(3) or - sqrt(3)/3
To confirm your answer, let's go through the steps of solving the problem together.
We are given that sin(theta) = -1/2 and theta is in quadrant IV. To find the cos and tan of theta, we can use the trigonometric identity:
sin^2(theta) + cos^2(theta) = 1.
Let's substitute the given value of sin(theta) into this equation:
(-1/2)^2 + cos^2(theta) = 1.
Simplifying, we get:
1/4 + cos^2(theta) = 1.
Subtracting 1/4 from both sides of the equation, we have:
cos^2(theta) = 3/4.
To solve for cos(theta), we can take the square root of both sides of the equation:
cos(theta) = √(3/4).
Now, since we know theta is in quadrant IV, where cosine is positive, the correct value for cos(theta) is positive √(3/4). However, note that √(3/4) can be simplified further:
cos(theta) = √(3/4) = √3/√4 = √3/2.
So, the correct value for cos(theta) is √3/2.
Finally, to find the tan(theta), we can use the identity:
tan(theta) = sin(theta)/cos(theta).
Substituting the given values, we have:
tan(theta) = (-1/2) / (√3/2).
Note that the denominator (√3/2) is the same as cos(theta), so we can simplify the expression:
tan(theta) = -1/√3.
To rationalize the denominator, we multiply both the numerator and denominator by √3:
tan(theta) = -√3 / 3.
Therefore, the correct values for cos(theta) and tan(theta) when sin(theta) is -1/2 in quadrant IV are:
cos(theta) = √3/2
tan(theta) = -√3/3
I hope this clarifies the correct answers for you!