Half angle identity

[IsaiahG0701]

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

[Joe Horn]

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

[Steve Simpkin]

As Joe pointed out. So many alternate forms for the answer.

[Albert Chan]

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)

[Albert Chan]

(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)