Half angle formulas

[VP1]

I have a question which states-
Suppose cotangent (x)= -3/5 where x is in the set of (3pi/2, 2 pi). find the exact value of sin(x/2)

So I'm pretty sure I'm supposed to use the half angle formulas.
I've figured out that cos(x)= 3/sqrt 34.

So the half angle formula for sin is- plus or minus sqrt [1-cos(x)/2]

Next I need to do the algebra. I just want to make sure I'm even using the right formula. Am I on the right track?

Thanks a lot- Todd

 

[soroban]

Hello, VP1!

$\displaystyle \cot x \,=\, \text{-}\tfrac{3}{5},\;x\text{ in Quadrant 4.}$
$\displaystyle \text{Find the exact value of }\sin\left(\tfrac{x}{2}\right)$

So I'm pretty sure I'm supposed to use the half angle formulas.
I've figured out that: cos x = 3/sqrt{34} . Yes!

So the half angle formula for sin is: plus or minus sqrt [(1-cos x)/2] . Right!

Next I need to do the algebra.
I just want to make sure I''m even using the right formula.
Am I on the right track? . Absolutely!

$\displaystyle \text{Note: if }x\text{ in in Quadrant 4, then }\tfrac{x}{2}\text{ is in Quadrant 3.}$

$\displaystyle \text{Select the sign of your answer accordingly.}$

 

[tkhunny]

Maybe right. The half-angle formula for the sine of THIS situation is NOT +/- anything. You are in Quadrant IV. The sine is negative.
Half of the angle is in Quadrant II. The sine is positive.

That formula is only a hint.

Edited for "What was I thinking?!"

 

[Deleted member 4993]

If x is in QIV (3?/2<x<2?) then x/2 is in QII

x = 2? - e ? x/2 = ? - e/2 (QII)

x = 3?/2 + e ? x/2 = 3?/4 + e/2 (QII)