Given tanu= -1/3 and sinu <0, find sin(u/2)
if 0 is less than or equal to u < 2 pie
I think its a right triangle in fourth quadrant with the point (3,-1) (hypotenuse equals square root of ten) but I'm brain dead and can't find u or any solutions.
I assume you want to solve for u, not a triangle. Tan u is negative in the second and fourth quadrant, and sie u is negative in the third and fourth, therefore you are correct that u must be in the fourth quadrant. The reference angle (which it makes with the +x axis is arctan 1/3.) You are correct that a line between (3, -1) and the origin will make an agle u with the x axis. That angle is 18.435 degrees and the angle u is 341.565 degrees
To find the value of sin(u/2), where tan(u) = -1/3 and sin(u) < 0, we can use the trigonometric half-angle identity for sine:
sin(u/2) = ±√[(1 - cos(u))/2]
Now, let's start by finding the value of cos(u):
cos(u) = 1 / √(1 + tan^2(u))
= 1 / √(1 + (-1/3)^2)
= 1 / √(1 + 1/9)
= 1 / √(10/9)
= √9 / √10
= 3 / √10
Since tan(u) = -1/3, we know that the tangent is negative in the fourth quadrant. From the given range of u (0 ≤ u < 2π), we can determine that u lies in the fourth quadrant.
In the fourth quadrant, both sine and cosine are negative. Therefore, sin(u) = -√(1 - cos^2(u))
= -√(1 - (3 / √10)^2)
= -√(1 - 9/10)
= -√(1/10)
= -1/√10
= -√10 / 10
Now we can substitute the value of sin(u) into the half-angle identity:
sin(u/2) = ±√[(1 - cos(u))/2]
= ±√[(1 - (3 / √10)) / 2]
= ±√[(√10 - 3) / (2√10)]
= ±(√10 - 3) / (2√10)
Since sin(u) < 0, we take the negative value of sin(u/2):
sin(u/2) = -(√10 - 3) / (2√10)
Therefore, sin(u/2) = -(√10 - 3) / (2√10) in the given range (0 ≤ u < 2π) where tan(u) = -1/3 and sin(u) < 0.