Determine Sign of Half-Angle Identities

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The discussion focuses on evaluating the half-angle identity for sine, specifically f(θ/2) given f(x) = sin(x). Participants address the calculation of y using the Pythagorean theorem and the correct application of the half-angle identity. There is confusion regarding the signs of the sine function based on the quadrant in which the angle lies. The correct approach involves using the x-coordinate for cos(θ) instead of the y-coordinate, leading to the simplified result of 3/√14. Understanding the quadrant's influence on the sign of the sine function is crucial for accurate calculations.
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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?
 

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rhoadsy74 said:
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?

rhoadsy74 said:
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.
 
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?
 
it is supposed to alpha.
 
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?
 
rhoadsy74 said:
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}
 
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 believe I did make a mistake somewhere, just not sure where.
 
rhoadsy74 said:
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:

rhoadsy74 said:
x2+y2=1

x= -2/7

(-2/7)2+y2=1

y=√(6)/7

Another thing:
rhoadsy74 said:
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?
 
  • #10
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?
 
  • #11
rhoadsy74 said:
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.

rhoadsy74 said:
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.
 
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