Question

If sin (B) = -1/3 with B in QIII, find csc (B/2).

Answer 1

\[\sin \left( \frac{B}{2} \right)=\sqrt{\frac{1-\cos B}{2}}\]

Answer 2

Right, but how does this help find csc?

Answer 3

\[\csc \left( \frac{B}{2} \right)=\frac{1}{\sin \left( \frac{B}{2} \right)}\]

Answer 4

Oh!!! I didn't think about it that way... So, if it's B/2, would I do (-1/3)/2?

Answer 5

no, it doesnt work that way, dividing the angle by 2 is not the same as dividing the value by 2

Answer 6

Good point. I'm sorry. This stuff is so confusing to me.

Answer 7

given sin B = -1/3, you need to find cos B first, and then plug it in into the half angle formula

Answer 8

Ok... So, to find cos B, I use the formula\[\pm \sqrt{(1+cosB)/2}\]

Answer 9

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