SOLUTION: it is given that sin A = K, where A is an angle measured in radians, and pi < A < (3pi/2) If sin B = k, which of the following is the value of B? A) A - pi B) pi + A C) 2

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: it is given that sin A = K, where A is an angle measured in radians, and pi < A < (3pi/2) If sin B = k, which of the following is the value of B? A) A - pi B) pi + A C) 2      Log On


   

Question 1135868: it is given that sin A = K, where A is an angle measured in radians,
and pi < A < (3pi/2)
If sin B = k, which of the following is the value of B?
A) A - pi
B) pi + A
C) 2pi - A
D) 3pi - A
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

The identities I will be using are
sin(A+B) = sin(A)cos(B) + cos(A)sin(B)
and
sin(A-B) = sin(A)cos(B) - cos(A)sin(B)

Let's the second identity mentioned to expand out choice A like so
sin(A - pi) = sin(A)cos(pi) - cos(A)sin(pi)
sin(A - pi) = sin(A)(-1) - cos(A)(0)
sin(A - pi) = -sin(A)
sin(A - pi) = -K
Choice A is crossed off the list because we want K instead of -K

Use the first identity to expand out the sine value of choice B
sin(pi+A) = sin(pi)cos(A) + cos(pi)sin(A)
sin(pi+A) = (0)cos(A) + (-1)sin(A)
sin(pi+A) = -sin(A)
sin(pi+A) = -K
we run into the same issue we did in choice A. So choice B is crossed off the list as well.

Go back to the second identity and expand for choice C
sin(2pi - A) = sin(2pi)cos(A) - cos(2pi)sin(A)
sin(2pi - A) = (0)cos(A) - (1)sin(A)
sin(2pi - A) = -sin(A)
sin(2pi - A) = -K
Same issue. Cross choice C off the list.

Repeat for choice D
sin(3pi - A) = sin(3pi)cos(A) - cos(3pi)sin(A)
sin(3pi - A) = (0)cos(A) - (-1)sin(A)
sin(3pi - A) = sin(A)
sin(3pi - A) = K
So this proves that sin(B) = K where B = 3pi-A. We have our answer

Answer: Choice D) 3pi - A