I'm not entirely sure how to determine whether the answer should be positive or negative (the quadrant, perhaps?), so I just left it there.
Anyways, thanks for your help. It's much more clear now.
sin(\frac{150^\circ}{2}) = \pm \sqrt{\frac{2+\sqrt{3}}{4}}
sin(\frac{150^\circ}{2}) = \pm...
Yes. As you stated, this problem can be solved easily with the sum formula for sin, but the directions ask us specifically to solve it using half angle identities. My problem is that I get stuck at that last step, as I'm not quite sure how to simplify it further.
Homework Statement
sin(75^\circ)
Homework Equations
sin(\frac{\alpha}{2}) = \pm \sqrt{\frac{1-cos(\alpha)}{2}}
The answer is known to be:
\frac{\sqrt{6} + \sqrt{2}}{4}
The Attempt at a Solution
sin(\frac{150^\circ}{2}) = \pm \sqrt{\frac{1-cos(150^\circ)}{2}}
sin(\frac{150^\circ}{2}) = \pm...