Determine Sign of Half-Angle Identities

[rhoadsy74]

Homework Statement

Use the figure to evaluate the function that f(x)=sin(x)

f(θ/2)

Homework Equations

n/a

The Attempt at a Solution

x²+y²=1

x= -2/7

(-2/7)²+y²=1

y=√(6)/7
sin(θ/2)= +/- √(1-cos(θ)/2) (the whole function is over 2 inside of the square root)

=+/- √(1-√(6)/7/2)

=+/- √(7-√(6)/14)

I keep coming out with the wrong sign. How do you determine the sign in a Half- Angle identity?

 

[eumyang]

x²+y²=1

x= -2/7

(-2/7)²+y²=1

y=√(6)/7

^ Not sure how you got that. Also, the y value has to be negative. Do you know why?

sin(θ/2)= +/- √(1-cos(θ)/2) (the whole function is over 2 inside of the square root)

=+/- √(1-√(6)/7/2)

=+/- √(7-√(6)/14)

I keep coming out with the wrong sign. How do you determine the sign in a Half- Angle identity?

Is it supposed to be theta or alpha? Your diagram has the angle labeled alpha. Assuming theta, and assuming you know what quadrant it is in, you can guess what quadrant θ/2 is in. You also should know that certain trig functions are positive in certain quadrants.

 

[rhoadsy74]

I used the Pythagorean theorem to find the (square root) 6/7 but it should be negative because of the fact that it is the third quadrant. What I don't understand is how to determine whether the whole function is (+) or (-). Can you help me?

 

[rhoadsy74]

it is supposed to alpha.

 

[rhoadsy74]

Why should it be negative? I understand that there are certain trig functions that are (+) and (-) in certain quadrants, but why is the the whole function negative?

 

[eumyang]

I used the Pythagorean theorem to find the (square root) 6/7 but it should be negative because of the fact that it is the third quadrant.

I'm afraid you made a mistake somewhere. Even with the negative, I am not getting
$$y = -\frac{\sqrt{6}}{7}$$

 

[rhoadsy74]

The answer to this question is 3/(square root) 14. That is what my homework is telling me. How do you come to that answer?

 

[rhoadsy74]

I believe I did make a mistake somewhere, just not sure where.

 

[eumyang]

The answer to this question is 3/(square root) 14. That is what my homework is telling me. How do you come to that answer?

I was not answering the overall question. I was pointing out that you got the wrong y-coordinate of the point in the beginning of your work:

x²+y²=1

x= -2/7

(-2/7)²+y²=1

y=√(6)/7

Another thing:

sin(θ/2)= +/- √(1-cos(θ)/2) (the whole function is over 2 inside of the square root)

=+/- √(1-√(6)/7/2)

I forgot to mention that this is also wrong. On the unit circle, does cos θ equal the y-coordinate?

 

[rhoadsy74]

Cos of theta equals the x- coordinate. I realize that I got it wrong, but can you explain to me where i went wrong and how the homework came to that answer?

 

[eumyang]

Cos of theta equals the x- coordinate. I realize that I got it wrong, but can you explain to me where i went wrong and how the homework came to that answer?

It would be better if you try it again yourself.

sin(θ/2)= +/- √(1-cos(θ)/2) (the whole function is over 2 inside of the square root)

=+/- √(1-√(6)/7/2)

Instead of plugging in the y-coordinate (which you still haven't fixed), plug in the x-coordinate, (-2/7):
$\sin \left ( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 - \cos \theta}{2}}$
$\sin \left ( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 - (-2/7)}{2}}$

This will simplify to $\frac{3}{\sqrt{14}}$. Try it.