Half-Angle Formula - Getting the Right Sign
Date: 06/11/2003 at 08:48:49 From: Prizzi Subject: Half-angle formulae: I can't get the sign right Find the possible values of tan(1/2)theta when tantheta = 7/24. I get the correct numerical answers 1/7(tan in the first quadrant) and 7(tan in the third quadrant), but for my answer when considering tan in the third quadrant, the answer at the back of my book gives -7. I used the half-angle formulae for cos^2(1/2)theta and sin^2(1/2) theta, substituting it into the trig identity sintheta/costheta = tantheta to find tan^2(1/2)theta, and taking the square root I got tan(1/2)theta.
Date: 06/11/2003 at 12:39:36 From: Doctor Peterson Subject: Re: Half-angle formulae: I can't get the sign right Hi, Prizzi. Our Formulas FAQ lists these half-angle identities: |sin(x/2)| = sqrt([1-cos(x)]/2), |cos(x/2)| = sqrt([1+cos(x)]/2), |tan(x/2)| = sqrt([1-cos(x)]/[1+cos(x)]), tan(x/2) = [1-cos(x)]/sin(x), = sin(x)/[1+cos(x)]. The absolute values on the first three indicate that you have to decide what sign to use based on the quadrant in which you expect the answer to lie. You apparently used these, dividing to get tan^2 before taking the square root, which is a good idea. But when you took the square root, you were not done, since there are TWO square roots of any positive number, and you had to choose one. I hope the version of the half-angle identities you used makes this clear. If theta is in the first quadrant, so is theta/2, and the tangent is positive. But if theta is in the third quadrant, then theta/2 is in the second quadrant, and you have to choose the negative tangent. If you used the last identity I showed above, you would not have a square root to find at the end; rather, you would make the choice of sign when you decided what the sine and cosine of theta itself are, before putting them into the formula. Try that method, and see that you get the right signs out if you put the right signs in. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 06/11/2003 at 14:36:16 From: Prizzi Subject: Thank you (Half-angle formulae: I can't get the sign right) Thank you for your help! I tried your version and finally have the correct signs. However I'm finding it difficult to follow how you derived tan(x/2) = [1-cos(x)]/sin(x),= sin(x)/[1+cos(x)] from |tan(x/2)| = sqrt([1-cos(x)]/[1+cosx)]). If it is not too much trouble, could you show me what you did? I'll probably be able to remember this formula if I understand how it is derived. Thank you!
Date: 06/11/2003 at 14:58:13 From: Doctor Peterson Subject: Re: Thank you (Half-angle formulae: I can't get the sign right) Hi, Prizzi. I didn't derive these formulas myself, but took them from the Dr. Math Trigonometry Formulas FAQ: http://mathforum.org/dr.math/faq/formulas/faq.trig.html#identities But let's see what we can do. We know that |tan(x/2)| = sqrt([1-cos(x)]/[1+cos(x)]) which we can write as 1-cos(x) tan(x/2) = +/- sqrt(--------) 1+cos(x) Multiplying numerator and denominator by 1-cos(x) gives [1-cos(x)][1-cos(x)] [1-cos(x)]^2 tan(x/2) = +/- sqrt(-------------------- = +/- sqrt(------------) [1+cos(x)][1-cos(x)] 1 - cos^2(x) 1 - cos(x) = +/- ---------- sin(x) All that is left is to choose the sign. Since tan(x/2) is positive when x/2 is in the first or third quadrant, it is positive when x is in the first or second quadrant, that is, when sin(x) > 0. That means we can just drop the "+/-", since 1-cos(x) > 0 and we just need the sign of the sine. You can derive the other version the same way. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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