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[quote="FrediFizzx"][quote="Yablon"]
OK, what I derived from half angle formulas for sin, cos and tan, via cos^2 and sin^2, is the following:

[tex]\frac{1}{\sqrt{2}}\sqrt{1+\cos \varphi \sin \theta }=\frac{1}{\sqrt{2}}\sqrt{1+\left( 4{{\cos }^{2}}\frac{\varphi }{2}-2 \right)\sin \frac{\theta }{2}\cos \frac{\theta }{2}}[/tex]

[tex]\frac{1}{\sqrt{2}}\sqrt{1+\sin \varphi \sin \theta }=\frac{1}{\sqrt{2}}\sqrt{1+\left( 4\sin \frac{\varphi }{2}\cos \frac{\varphi }{2} \right)\sin \frac{\theta }{2}\cos \frac{\theta }{2}}[/tex]

I would have to say that the original expressions in this case are simpler than those with half angles. Anybody see anything else?

Jay[/quote]
Mathematica gives for the first one,

[tex]\text{TrigReduce}\left[\frac{\sqrt{\sin (\theta ) \cos (\phi )+1}}{\sqrt{2}}\right]=\frac{1}{2} \sqrt{\sin (\theta -\phi )+\sin (\theta +\phi )+2}[/tex]

And the second one,

[tex]\text{TrigReduce}\left[\frac{\sqrt{\sin (\theta ) \sin (\phi )+1}}{\sqrt{2}}\right]=\frac{1}{2} \sqrt{\cos (\theta -\phi )-\cos (\theta +\phi )+2}[/tex]
.[/quote]
Here is a spherical plot of the first one. The side you can't see is just a round sphere.

[img]http://www.sciphysicsforums.com/spfbb1/EPRsims/sphericalplot.jpg[/img]
.

[quote="Yablon"][quote="Yablon"]Hi all:

You will all know the trigonometric half-angle identity [tex]\cos \frac{\alpha }{2}=+\frac{1}{\sqrt{2}}\sqrt{1+\cos \alpha }[/tex] for [tex]0 \le \alpha \le \pi[/tex].

Can someone please figure out, or does someone know, something similar for the following?

[tex]???=\frac{1}{\sqrt{2}}\sqrt{1+\cos \varphi \sin \theta }[/tex]

[tex]???=\frac{1}{\sqrt{2}}\sqrt{1+\sin \varphi \sin \theta }[/tex]

Either point to a trigonometry reference, or please show us how you get there. I am concerned about the body of the functions, more so than the signs.

Thanks,

Jay[/quote]
OK, what I derived from half angle formulas for sin, cos and tan, via cos^2 and sin^2, is the following:

[tex]\frac{1}{\sqrt{2}}\sqrt{1+\cos \varphi \sin \theta }=\frac{1}{\sqrt{2}}\sqrt{1+\left( 4{{\cos }^{2}}\frac{\varphi }{2}-2 \right)\sin \frac{\theta }{2}\cos \frac{\theta }{2}}[/tex]

[tex]\frac{1}{\sqrt{2}}\sqrt{1+\sin \varphi \sin \theta }=\frac{1}{\sqrt{2}}\sqrt{1+\left( 4\sin \frac{\varphi }{2}\cos \frac{\varphi }{2} \right)\sin \frac{\theta }{2}\cos \frac{\theta }{2}}[/tex]

I would have to say that the original expressions in this case are simpler than those with half angles. Anybody see anything else?

Jay[/quote]
Mathematica gives for the first one,

[tex]\text{TrigReduce}\left[\frac{\sqrt{\sin (\theta ) \cos (\phi )+1}}{\sqrt{2}}\right]=\frac{1}{2} \sqrt{\sin (\theta -\phi )+\sin (\theta +\phi )+2}[/tex]

And the second one,

[tex]\text{TrigReduce}\left[\frac{\sqrt{\sin (\theta ) \sin (\phi )+1}}{\sqrt{2}}\right]=\frac{1}{2} \sqrt{\cos (\theta -\phi )-\cos (\theta +\phi )+2}[/tex]
.

[quote="Yablon"]Hi all:

You will all know the trigonometric half-angle identity [tex]\cos \frac{\alpha }{2}=+\frac{1}{\sqrt{2}}\sqrt{1+\cos \alpha }[/tex] for [tex]0 \le \alpha \le \pi[/tex].

Can someone please figure out, or does someone know, something similar for the following?

[tex]???=\frac{1}{\sqrt{2}}\sqrt{1+\cos \varphi \sin \theta }[/tex]

[tex]???=\frac{1}{\sqrt{2}}\sqrt{1+\sin \varphi \sin \theta }[/tex]

Either point to a trigonometry reference, or please show us how you get there. I am concerned about the body of the functions, more so than the signs.

Thanks,

Jay[/quote]
OK, what I derived from half angle formulas for sin, cos and tan, via cos^2 and sin^2, is the following:

[tex]\frac{1}{\sqrt{2}}\sqrt{1+\cos \varphi \sin \theta }=\frac{1}{\sqrt{2}}\sqrt{1+\left( 4{{\cos }^{2}}\frac{\varphi }{2}-2 \right)\sin \frac{\theta }{2}\cos \frac{\theta }{2}}[/tex]

[tex]\frac{1}{\sqrt{2}}\sqrt{1+\sin \varphi \sin \theta }=\frac{1}{\sqrt{2}}\sqrt{1+\left( 4\sin \frac{\varphi }{2}\cos \frac{\varphi }{2} \right)\sin \frac{\theta }{2}\cos \frac{\theta }{2}}[/tex]

I would have to say that the original expressions in this case are simpler than those with half angles. Anybody see anything else?

Jay

Hi all:

You will all know the trigonometric half-angle identity [tex]\cos \frac{\alpha }{2}=+\frac{1}{\sqrt{2}}\sqrt{1+\cos \alpha }[/tex] for [tex]0 \le \alpha \le \pi[/tex].

Can someone please figure out, or does someone know, something similar for the following?

[tex]???=\frac{1}{\sqrt{2}}\sqrt{1+\cos \varphi \sin \theta }[/tex]

[tex]???=\frac{1}{\sqrt{2}}\sqrt{1+\sin \varphi \sin \theta }[/tex]

Either point to a trigonometry reference, or please show us how you get there. I am concerned about the body of the functions, more so than the signs.

Thanks,

Jay