Half angle identity - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: HP Prime (/forum-5.html) +--- Thread: Half angle identity (/thread-18270.html) |
Half angle identity - IsaiahG0701 - 04-17-2022 12:51 AM I’m struggling with half angle identity problems. For example, when I compute Cos (105°) I get -6 root + 2 root / 4. But the actual answer is -2 root - 3 root / 2. Is there a reason it is not simplifying to this number or would it be a user error RE: Half angle identity - Joe Horn - 04-17-2022 03:43 AM If you mean that the correct answer is \(\cfrac{-\sqrt{2-\sqrt{3}}}{2}\), you're right, but so is Prime's answer. They're equivalent ways of expressing the same number, approximately -0.258819045103... If you didn't mean the above expression, compare with the output from https://www.wolframalpha.com/input?i=cos%28105%C2%B0%29 RE: Half angle identity - Steve Simpkin - 04-17-2022 07:18 AM As Joe pointed out. So many alternate forms for the answer. RE: Half angle identity - Albert Chan - 04-17-2022 03:29 PM cos(105°) = sin(90-105°) = -sin(15°) = -√((1-cos(30°)/2) = -√((1-√3/2)/2) We can remove nested square roots with identity (easily confirmed by squaring both side) If x,y square root free, RHS (assumed ≥ 0) have no nested square roots. \(\sqrt{2\;(x ± \sqrt{x^2-y^2})} = \sqrt{x+y}\;± \sqrt{x-y} \) -√((1-√3/2)/2) = -1/2 * √(2*(1 - √(1-1/4))) = -1/2 * (√(1+1/2) - √(1-1/2)) = (-√6 + √2)/4 = -1/(√6 + √2) RE: Half angle identity - Albert Chan - 04-17-2022 05:09 PM (04-17-2022 03:29 PM)Albert Chan Wrote: \(\sqrt{2\;(x ± \sqrt{x^2-y^2})} = \sqrt{x+y}\;± \sqrt{x-y} \) We can use the identity to build formula for complex square roots Let Z = X+Y*i. For simplify assume Z on the unit circle. \(\displaystyle \sqrt{2\;(X ± 1)} = \sqrt{Z}\;± \sqrt{\bar{Z}} \) \(\displaystyle \sqrt{2Z} = \sqrt{1+X} + i \; sgn(Y)\; \sqrt{1-X} = \frac{(1+X)\;+\;i\;Y}{\sqrt{1+X}} \) \(\displaystyle \;\,\sqrt{Z} = \frac{Z+1}{|Z+1|}\) // if |Z| = 1 Let Z = cis(θ). Matching parts of √Z = cis(θ/2), above gives half-angle formulas: Assume θ = arg(z), with range ± pi cos(θ/2) = (1+cos(θ)) / √(2+2 cos(θ)) = √((1+cos(θ))/2) ≥ 0 sin(θ/2) = sin(θ) / √(2+2 cos(θ)) = √((1−cos(θ))/2) * sgn(θ) |