assume that all the given angles are in simplest form, so that if A is in QIV you may assume that 270° < A < 360°. If sin B = − 1/3 with B in QIII, find the following. sin B/2

Answer 1

Hey! Lets solve this shall we ?

So, we must find out what sin(B/2) is given sinB = -1/3 and 270^o < A < 360^o . First, we solve for B such that we take the inverse sine function of both sides such that

sin^-1sinB = sin^-1(-1/3)

B = sin^-1(-1/3)

B = -19.47 (degrees)

Then, we take that to find sin(-19.47/2) using the half-angle formula such that

sin(B/2) = √[1-cos(-19.47)/2] = √[.0571838573/2] = √[.0285919286] = -0.169

I hope this helped!

Answer 2

cosB = -√[1-sin²B] = -√[1-1/9] = -√8/3. (Sine and cosine are both negative in Quadrant 3).

Since B is in Quadrant 3, 180<B<270. So, 90<B/2<135. Therefore, B/2 lies in Quadrant 2 so sin(B/2)>0.

Using the half-angle formula for sine, we have sin(B/2) = √[(1-cosB)/2] = √[(1+√8/3)/2]

Answer 3

asin-.333..........=-19.469..............°

quad3=199.46.................°=b

sin(199.46......./2)=.985.............°