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[FloridaDean]

understand wemen is like:

displaystyle Substitute; x=cosalp, alpin[0: ;: pi ]Substitutex=cosalp,alp∈[0:;π]
displaystyle Then; |x+sqrt{1-x^2}|=sqrt{2}(2x^2-1)Leftright |cosalp +sinalp |=sqrt{2}(2cos^2alp -1)Then∣x+
1−x 2∣= 2(2x 2 −1)Leftright∣cosalp+sinalp∣=
2(2cos
2alp−1)
displaystyle |N {sqrt{2}}cos(alp-frac{pi}{4})|=N {sqrt{2}}cos(2alp )Right alpin[0: ;: frac{pi}{4}]cup [frac{3pi}{4}: ;: pi]∣N
2cos(alp−
4π )∣=N 2cos(2alp)Rightalp∈[0;
4π]∪[ 43π;π]1) displaystyle alp in [0: ;: frac{pi}{4}]alp∈[0;
4π]displaystyle cos(alp -frac{pi}{4})=cos(2alp )dotscos(alp−
4π )=cos(2alp)…
2. displaystyle alpin [frac{3pi}{4}: ;: pi]alp∈[
43π ;π]displaystyle -cos(alp -frac{pi}{4})=cos(2alp )dots−cos(alp−
4π
 )=cos(2alp)…
1

That makes perfect sense.

For a moment there I thought we had another troll visit.

spellcheck shut down on me.

[Threebean]

understand wemen is like:

displaystyle Substitute; x=cosalp, alpin[0: ;: pi ]Substitutex=cosalp,alp∈[0:;π]
displaystyle Then; |x+sqrt{1-x^2}|=sqrt{2}(2x^2-1)Leftright |cosalp +sinalp |=sqrt{2}(2cos^2alp -1)Then∣x+
1−x 2∣= 2(2x 2 −1)Leftright∣cosalp+sinalp∣=
2(2cos
2alp−1)
displaystyle |N {sqrt{2}}cos(alp-frac{pi}{4})|=N {sqrt{2}}cos(2alp )Right alpin[0: ;: frac{pi}{4}]cup [frac{3pi}{4}: ;: pi]∣N
2cos(alp−
4π )∣=N 2cos(2alp)Rightalp∈[0;
4π]∪[ 43π;π]1) displaystyle alp in [0: ;: frac{pi}{4}]alp∈[0;
4π]displaystyle cos(alp -frac{pi}{4})=cos(2alp )dotscos(alp−
4π )=cos(2alp)…
2. displaystyle alpin [frac{3pi}{4}: ;: pi]alp∈[
43π ;π]displaystyle -cos(alp -frac{pi}{4})=cos(2alp )dots−cos(alp−
4π
 )=cos(2alp)…
1

That makes perfect sense.

For a moment there I thought we had another troll visit.

spellcheck shut down on me.

That post makes it look like sanity check shutdown on you, too.

[A Friend of Charlie]

understand wemen is like:

displaystyle Substitute; x=cosalp, alpin[0: ;: pi ]Substitutex=cosalp,alp∈[0:;π]
displaystyle Then; |x+sqrt{1-x^2}|=sqrt{2}(2x^2-1)Leftright |cosalp +sinalp |=sqrt{2}(2cos^2alp -1)Then∣x+
1−x 2∣= 2(2x 2 −1)Leftright∣cosalp+sinalp∣=
2(2cos
2alp−1)
displaystyle |N {sqrt{2}}cos(alp-frac{pi}{4})|=N {sqrt{2}}cos(2alp )Right alpin[0: ;: frac{pi}{4}]cup [frac{3pi}{4}: ;: pi]∣N
2cos(alp−
4π )∣=N 2cos(2alp)Rightalp∈[0;
4π]∪[ 43π;π]1) displaystyle alp in [0: ;: frac{pi}{4}]alp∈[0;
4π]displaystyle cos(alp -frac{pi}{4})=cos(2alp )dotscos(alp−
4π )=cos(2alp)…
2. displaystyle alpin [frac{3pi}{4}: ;: pi]alp∈[
43π ;π]displaystyle -cos(alp -frac{pi}{4})=cos(2alp )dots−cos(alp−
4π
 )=cos(2alp)…
1

That makes perfect sense.

For a moment there I thought we had another troll visit.

spellcheck shut down on me.

That post makes it look like sanity check shutdown on you, too.

Damn, is it Tuesday already?

[FloridaDean]

understand wemen is like:

displaystyle Substitute; x=cosalp, alpin[0: ;: pi ]Substitutex=cosalp,alp∈[0:;π]
displaystyle Then; |x+sqrt{1-x^2}|=sqrt{2}(2x^2-1)Leftright |cosalp +sinalp |=sqrt{2}(2cos^2alp -1)Then∣x+
1−x 2∣= 2(2x 2 −1)Leftright∣cosalp+sinalp∣=
2(2cos
2alp−1)
displaystyle |N {sqrt{2}}cos(alp-frac{pi}{4})|=N {sqrt{2}}cos(2alp )Right alpin[0: ;: frac{pi}{4}]cup [frac{3pi}{4}: ;: pi]∣N
2cos(alp−
4π )∣=N 2cos(2alp)Rightalp∈[0;
4π]∪[ 43π;π]1) displaystyle alp in [0: ;: frac{pi}{4}]alp∈[0;
4π]displaystyle cos(alp -frac{pi}{4})=cos(2alp )dotscos(alp−
4π )=cos(2alp)…
2. displaystyle alpin [frac{3pi}{4}: ;: pi]alp∈[
43π ;π]displaystyle -cos(alp -frac{pi}{4})=cos(2alp )dots−cos(alp−
4π
 )=cos(2alp)…
1

That makes perfect sense.

For a moment there I thought we had another troll visit.

spellcheck shut down on me.

That post makes it look like sanity check shutdown on you, too.

Fish math.

[Threebean]

Likely my last cigar until the weekend so might as well make it a good one, with coffee.

[FloridaDean]

rule #1: the woman is always right.
rule #2: see rule #1.

[FloridaDean]

Likely my last cigar until the weekend so might as well make it a good one, with coffee.

everyday is Monday, or neverending weekend.

[bluecollar]

understand wemen is like:

displaystyle Substitute; x=cosalp, alpin[0: ;: pi ]Substitutex=cosalp,alp∈[0:;π]
displaystyle Then; |x+sqrt{1-x^2}|=sqrt{2}(2x^2-1)Leftright |cosalp +sinalp |=sqrt{2}(2cos^2alp -1)Then∣x+
1−x 2∣= 2(2x 2 −1)Leftright∣cosalp+sinalp∣=
2(2cos
2alp−1)
displaystyle |N {sqrt{2}}cos(alp-frac{pi}{4})|=N {sqrt{2}}cos(2alp )Right alpin[0: ;: frac{pi}{4}]cup [frac{3pi}{4}: ;: pi]∣N
2cos(alp−
4π )∣=N 2cos(2alp)Rightalp∈[0;
4π]∪[ 43π;π]1) displaystyle alp in [0: ;: frac{pi}{4}]alp∈[0;
4π]displaystyle cos(alp -frac{pi}{4})=cos(2alp )dotscos(alp−
4π )=cos(2alp)…
2. displaystyle alpin [frac{3pi}{4}: ;: pi]alp∈[
43π ;π]displaystyle -cos(alp -frac{pi}{4})=cos(2alp )dots−cos(alp−
4π
 )=cos(2alp)…
1

...but how can you read it Dean?..

[bluecollar]

Likely my last cigar until the weekend so might as well make it a good one, with coffee.

everyday is Monday, or neverending weekend.

I like the Mexican habano wrapper. Good cigar. Been a while on that one. How did you like it?

[Travellin Dave]

understand wemen is like:

displaystyle Substitute; x=cosalp, alpin[0: ;: pi ]Substitutex=cosalp,alp∈[0:;π]
displaystyle Then; |x+sqrt{1-x^2}|=sqrt{2}(2x^2-1)Leftright |cosalp +sinalp |=sqrt{2}(2cos^2alp -1)Then∣x+
1−x 2∣= 2(2x 2 −1)Leftright∣cosalp+sinalp∣=
2(2cos
2alp−1)
displaystyle |N {sqrt{2}}cos(alp-frac{pi}{4})|=N {sqrt{2}}cos(2alp )Right alpin[0: ;: frac{pi}{4}]cup [frac{3pi}{4}: ;: pi]∣N
2cos(alp−
4π )∣=N 2cos(2alp)Rightalp∈[0;
4π]∪[ 43π;π]1) displaystyle alp in [0: ;: frac{pi}{4}]alp∈[0;
4π]displaystyle cos(alp -frac{pi}{4})=cos(2alp )dotscos(alp−
4π )=cos(2alp)…
2. displaystyle alpin [frac{3pi}{4}: ;: pi]alp∈[
43π ;π]displaystyle -cos(alp -frac{pi}{4})=cos(2alp )dots−cos(alp−
4π
 )=cos(2alp)…
1

That makes perfect sense.

For a moment there I thought we had another troll visit.

Maybe we did.

Lot of Russians in Florida.

[bluecollar]

On the Monte. May have miss quoted. Smoking an oliva Connecticut churchie while doing yard work

[Travellin Dave]

Still haven't found my ashtrays.

[A Friend of Charlie]

Likely my last cigar until the weekend so might as well make it a good one, with coffee.

It's a beautiful afternoon here but I don't think I'll be able to sneak in that cigar. I'm so frustrated.

[Threebean]

Likely my last cigar until the weekend so might as well make it a good one, with coffee.

everyday is Monday, or neverending weekend.

I like the Mexican habano wrapper. Good cigar. Been a while on that one. How did you like it?

Currently it's my favorite smoke. 

[Threebean]

Still haven't found my ashtrays.

Just use that toilet looking thing in the background.