Half angle identity
04-17-2022, 12:51 AM
Post: #1
 IsaiahG0701 Junior Member Posts: 9 Joined: Mar 2022
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
04-17-2022, 03:43 AM (This post was last modified: 04-17-2022 03:21 PM by Joe Horn.)
Post: #2
 Joe Horn Senior Member Posts: 2,005 Joined: Dec 2013
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

<0|ɸ|0>
-Joe-
04-17-2022, 07:18 AM (This post was last modified: 04-17-2022 03:35 PM by Steve Simpkin.)
Post: #3
 Steve Simpkin Senior Member Posts: 1,215 Joined: Dec 2013
RE: Half angle identity
As Joe pointed out. So many alternate forms for the answer.

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04-17-2022, 03:29 PM (This post was last modified: 04-17-2022 05:49 PM by Albert Chan.)
Post: #4
 Albert Chan Senior Member Posts: 2,682 Joined: Jul 2018
RE: Half angle identity
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)
04-17-2022, 05:09 PM
Post: #5
 Albert Chan Senior Member Posts: 2,682 Joined: Jul 2018
RE: Half angle identity
(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)
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