Half angle identity

04172022, 12:51 AM
Post: #1




Half angle identity
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


04172022, 03:43 AM
(This post was last modified: 04172022 03:21 PM by Joe Horn.)
Post: #2




RE: Half angle identity
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
<0ɸ0> Joe 

04172022, 07:18 AM
(This post was last modified: 04172022 03:35 PM by Steve Simpkin.)
Post: #3




RE: Half angle identity
As Joe pointed out. So many alternate forms for the answer.


04172022, 03:29 PM
(This post was last modified: 04172022 05:49 PM by Albert Chan.)
Post: #4




RE: Half angle identity
cos(105°) = sin(90105°) = sin(15°) = √((1cos(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. \(\sqrt{2\;(x ± \sqrt{x^2y^2})} = \sqrt{x+y}\;± \sqrt{xy} \) √((1√3/2)/2) = 1/2 * √(2*(1  √(11/4))) = 1/2 * (√(1+1/2)  √(11/2)) = (√6 + √2)/4 = 1/(√6 + √2) 

04172022, 05:09 PM
Post: #5




RE: Half angle identity
(04172022 03:29 PM)Albert Chan Wrote: \(\sqrt{2\;(x ± \sqrt{x^2y^2})} = \sqrt{x+y}\;± \sqrt{xy} \) 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. \(\displaystyle \sqrt{2\;(X ± 1)} = \sqrt{Z}\;± \sqrt{\bar{Z}} \) \(\displaystyle \sqrt{2Z} = \sqrt{1+X} + i \; sgn(Y)\; \sqrt{1X} = \frac{(1+X)\;+\;i\;Y}{\sqrt{1+X}} \) \(\displaystyle \;\,\sqrt{Z} = \frac{Z+1}{Z+1}\) // if Z = 1 Let Z = cis(θ). Matching parts of √Z = cis(θ/2), above gives halfangle 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) 

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