Triginometry: Half Angle Identities

[StephenAA]

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

$$sin(\frac{150^\circ}{2}) = \pm \sqrt{\frac{1-cos(150^\circ)}{2}}$$
$$sin(\frac{150^\circ}{2}) = \pm \sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}$$
$$sin(\frac{150^\circ}{2}) = \pm \sqrt{\frac{2+\sqrt{3}}{4}}$$
$$sin(\frac{150^\circ}{2}) = \pm \frac{\sqrt{2+\sqrt{3}}}{2}$$

This is a correct answer, but clearly it is not simplified completely. Any help would be appreciated.

 

[Ivan92]

Are you supposed to use the half-angle identity? Think of another way to figure it out.

Hint: Sum Formula for sin function.

 

[StephenAA]

Are you supposed to use the half-angle identity? Think of another way to figure it out.

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.

 

[eumyang]

First off, do we NEED the plus-minus sign? (If not, can you explain why not?)

The 2nd to last line you have this:
$$sin(\frac{150^\circ}{2}) = \pm \sqrt{\frac{2+\sqrt{3}}{4}}$$

Inside the square root, multiply top and bottom by 4. Factor out a 2 in the numerator (not 4!). You should have
$$\sin \left(\frac{150^\circ}{2}\right) = \sqrt{\frac{2(4 + 2\sqrt{3})}{16}}$$
Split the 4 into 1 + 3:
$$\sin \left(\frac{150^\circ}{2}\right) = \sqrt{\frac{2(1 + 2\sqrt{3} + 3)}{16}}$$

This expression:
$$1 + 2\sqrt{3} + 3$$
is an algebraic pattern that's somewhat disguised. See if you can figure out the rest.

 

[StephenAA]

First off, do we NEED the plus-minus sign? (If not, can you explain why not?)

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.
$$sin(\frac{150^\circ}{2}) = \pm \sqrt{\frac{2+\sqrt{3}}{4}}$$
$$sin(\frac{150^\circ}{2}) = \pm \sqrt{\frac{8+4\sqrt{3}}{16}}$$
$$sin(\frac{150^\circ}{2}) = \pm \sqrt{\frac{2(4+2\sqrt{3})}{16}}$$
$$sin(\frac{150^\circ}{2}) = \pm \sqrt{\frac{2(1+2\sqrt{3}+3)}{16}}$$
$$sin(\frac{150^\circ}{2}) = \pm \sqrt{\frac{2(1+\sqrt{3})^2}{16}}$$
$$sin(\frac{150^\circ}{2}) = \pm \frac{(\sqrt{2})(1+\sqrt{3})}{4}$$

For the final answer:
$$sin(75^\circ) = \pm \frac{\sqrt{2}+\sqrt{6}}{4}$$

Or, as the answer key lists it:
$$sin(75^\circ) = \pm \frac{\sqrt{6}+\sqrt{2}}{4}$$

Thank you for your help.

 

[Mentallic]

Yes, but not quite. sin(75^o) is a positive value, so you need to discard the negative value.