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+--- 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)