TRIG 1/2 ANGLE FORUMLAS

Question

The questions are: "If sin B = -1/3 with B in QIII, find tan B/2. Also find sin B/2

 

The answer for tan B/2 is-3-2sqrt(2)

 

The answer for sin B/2 is \sqrt((3+2\sqrt(2))/(6))

 

These are answers from the back of the book. I can't figure out how to work these problems out. Please help.

Answer 1

Hi Michael,

 

Think of the reference angle and associated triangle for angle B. Its opposite side will be -1 with a hypotenuse of 3. Use Pythagorean theorem to get adjacent side: 2√2 (which will be negative in QIII). This means the cos B = -2√2/3.

 

There are several different formulas for the tan B/2, one of them is (1-cos B)/sin B.

 

Substituting what we know:  (1 - (-2√2/3))/-1/3  gives  (1+2√2/3)/-1/3. This is a complex fraction which I would simplify by multiplying both numerator and denominator of the "big" fraction by 3.

 

(3+2√2)/-1  = -3-2√2

 

That is just one possible form for the answer. If you use tan B/2 = √(1-cosB)/(1+cosB) your answer would look different, but it would evaluate to the same value.