Quote from: A Friend of Charlie on October 21, 2019, 02:39:30 PMQuote from: FloridaDean on October 21, 2019, 02:32:40 PMunderstand 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)…1That makes perfect sense.For a moment there I thought we had another troll visit.
Quote from: FloridaDean on October 21, 2019, 02:32:40 PMunderstand 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)…1That makes perfect sense.
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
Quote from: Threebean on October 21, 2019, 02:44:21 PMQuote from: A Friend of Charlie on October 21, 2019, 02:39:30 PMQuote from: FloridaDean on October 21, 2019, 02:32:40 PMunderstand 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)…1That makes perfect sense.For a moment there I thought we had another troll visit.spellcheck shut down on me.
Quote from: FloridaDean on October 21, 2019, 02:58:34 PMQuote from: Threebean on October 21, 2019, 02:44:21 PMQuote from: A Friend of Charlie on October 21, 2019, 02:39:30 PMQuote from: FloridaDean on October 21, 2019, 02:32:40 PMunderstand 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)…1That 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.
Likely my last cigar until the weekend so might as well make it a good one, with coffee.
Quote from: Threebean on October 21, 2019, 04:14:26 PMLikely my last cigar until the weekend so might as well make it a good one, with coffee. everyday is Monday, or neverending weekend.
Quote from: Threebean on October 21, 2019, 02:44:21 PMQuote from: A Friend of Charlie on October 21, 2019, 02:39:30 PMQuote from: FloridaDean on October 21, 2019, 02:32:40 PMunderstand 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)…1That makes perfect sense.For a moment there I thought we had another troll visit.Maybe we did.
Quote from: FloridaDean on October 21, 2019, 04:16:27 PMQuote from: Threebean on October 21, 2019, 04:14:26 PMLikely 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?
Still haven't found my ashtrays.