sec theta = 2, sin theta is less than 0. I'm not sure what to do? Please explain. Thank you.
secTheta=2 means cosTheta=1/2, which means that Theta is 60 degrees.
Are you trying to find theta ?
if sec theta = 2 then cos theta = 1/2
the cosine is positive in the first and fourth quadrants.
We are also told that sin(theta) is negative, so theta must be either in the third or fourth quadrants.
Combininig the two conditions, we can say that theta is in the fourth quadrant.
since cos(theta) = 1/2 we have a reference angle of 60º
so theta = 360-60 = 300º or 5pi/3 radians
To solve the given equation sec(theta) = 2 and sin(theta) < 0, let's break it down step by step:
1. Recall the definitions:
- sec(theta) = 1 / cos(theta)
- sin(theta) = opposite/hypotenuse
2. Since sec(theta) = 2, we can rewrite it as 1 / cos(theta) = 2. To solve for cos(theta), we take the reciprocal of both sides:
cos(theta) = 1 / 2
3. To determine the sign of sin(theta), we'll need to find the quadrant in which theta lies, as sin(theta) is negative in quadrants III and IV.
4. Since sec(theta) = 2, which is greater than 1, we know that theta is an acute angle. However, as sec(theta) is positive, we can eliminate quadrant III, as sec(theta) is negative in that quadrant (thus not equal to 2).
5. Therefore, theta must lie in quadrant IV since sec(theta) > 1 and sec(theta) = 2. In this quadrant, cos(theta) is positive.
6. By using the Pythagorean identity (sin^2(theta) + cos^2(theta) = 1), we can substitute the value of cos(theta) we found in step 2:
sin^2(theta) + (1/2)^2 = 1
sin^2(theta) + 1/4 = 1
sin^2(theta) = 3/4
7. Taking the square root of both sides, we get:
sin(theta) = ±√(3/4)
With sin(theta) < 0, we only consider the negative square root:
sin(theta) = -√3/2
So, for sec(theta) = 2 and sin(theta) < 0, theta lies in quadrant IV and the values of theta that satisfy these conditions are:
- Theta = atan(-√3/2) + nπ, where n is an integer.
Note: atan refers to the inverse tangent function, which returns the angle whose tangent is a given value.