用户:Weichi6206/沙盒'

$\ \nu$

证明

证明 $\ tan{\frac {\theta }{2}}={\frac {1+sin\theta -cos\theta }{1+sin\theta +cos\theta }}$

右式 $\ ={\frac {1+sin\theta -cos\theta }{1+sin\theta +cos\theta }}$

上下同乘(1+sinθ-cosθ)/(1+sinθ-cosθ)

$\ ={\frac {1+sin\theta -cos\theta }{1+sin\theta +cos\theta }}\times {\frac {1+sin\theta -cos\theta }{1+sin\theta -cos\theta }}$

将1+sinθ一国，cosθ一国

$\ ={\frac {(1+sin\theta )^{2}-2cos\theta (1+sin\theta )+cos^{2}\theta }{(1+sin\theta )^{2}-cos^{2}\theta }}$

将 $cos^{2}\theta$ 换成 $1-sin^{2}\theta$

$\ ={\frac {(1+sin\theta )^{2}-2cos\theta (1+sin\theta )+(1-sin^{2}\theta )}{(1+sin\theta )^{2}-(1-sin^{2}\theta )}}$

将 $1-sin^{2}\theta$ 拆开成 $(1-sin\theta )(1+sin\theta )$

$\ ={\frac {(1+sin\theta )^{2}-2cos\theta (1+sin\theta )+(1-sin\theta )(1+sin\theta )}{(1+sin\theta )^{2}-(1-sin\theta )(1+sin\theta )}}$

将1+sinθ约掉

$\ ={\frac {(1+sin\theta )-2cos\theta +(1-sin\theta )}{(1+sin\theta )-(1-sin\theta )}}$

经过妥善的运算之后(某人口头禅)

$\ ={\frac {1-cos\theta }{sin\theta }}$

其实这就是 $tan{\frac {\theta }{2}}$

$\ tan{\frac {\theta }{2}}={\sqrt {\frac {1-cos\theta }{1+cos\theta }}}={\frac {1-cos\theta }{sin\theta }}={\frac {sin\theta }{1+cos\theta }}$

棒球术语

投手

earned run average 自责分率

$\ ERA={\frac {ER\times 9}{IP}}$

Hits per Nine Innings 被安打率

$\ H/9={\frac {H\times 9}{IP}}$

Walks Plus Hits per Inning Pitched 每局被上垒率

$\ WHIP={\frac {H+BB}{IP}}$

Batting Average on Balls put Into Play 场内被安打率

$\ BABIP={\frac {H-HR}{AB-SO-HR}}$

Walks per Nine Innings 保送率

$\ BB/9={\frac {BB+IBB+HBP\times 9}{IP}}$

Opponent On-base Plus Slugging 被攻击指数

$\ Oops=OBP+SLG$

Inning per Game Started 平均先发局数

$\ I/GS={\frac {IP}{GS}}$

打者

batting average 打击率

$\ AVG={\frac {H}{AB}}$

on base percentage 上垒率

$\ OBP={\frac {H+BB+IBB+HBP}{AB+BB+IBB+HBP+SF}}$

Total Bases On Safe Hits 垒打数

$\ TB=1\times 1B+2\times 2B+3\times 3B+4\times HR$

slugging percentage 长打率

$\ SLG={\frac {TB}{AB}}$

=Isolated Power 纯长打率=

$\ IsoP=SLG-AVG$

Balls put Into Play percentage 将球打进场内几率

$\ BIP\%={\frac {AB-SO-HR}{AB+BB+HBP+SF}}$

Isolated Discipline 选球率

$\ IsoD=OBP-AVG$

=Pure IsoD=

$\ IsoD=OBP-AVG-IBB$

On-base percentage Plus Slugging Percentage 整体攻击指数

$\ OPS=OBP+SLG$

=Total Average 攻击指数=

$\ TA={\frac {TB+SB+BB+IBB+HBP}{AB-H+CS+GIDP}}$

=Gross Production Average 整体攻击指数(平衡过的)=

$\ GPA={\frac {1.8\times OBP+SLG}{4}}$

probability of a successful steal 盗垒成功率

$\ PSS={\frac {SB}{SB+CS}}$

Offense Ratio 攻击比

$\ OR={\frac {TB+BB}{AB-H}}$

Range Factor 球员9局制造出局数

$\ RF={\frac {PO+A\times 9}{INN}}$

皆适用

Batting Park Factor 球场校正值

$\ BPF={\frac {\frac {homeRS+homeRA}{homeG}}{\frac {roadRS+homeRA}{homeG}}}$

逼近0

$\ k=\lim _{G\to 0}G\%$

金勋奖

现在*3 $\ k'=\lim _{G\to 0}3G\%$

PVC聚氯乙烯

乙烯与氯气在三氯化铁催化下加成生成1,2-二氯乙烷

$\ CH_{2}=CH_{2}+Cl_{2}\to ClCH_{2}CH_{2}Cl$

精制后的1,2-二氯乙烷发生裂解，生成氯乙烯和氯化氢

$\ ClCH_{2}CH_{2}Cl\to CH_{2}=CHCl+HCl$

由氯乙烯合成聚氯乙烯

$\ CH_{2}=CHCl\to (CH_{2}-CHCl)_{n}$

未知化合物

$\ CH_{3}COONH_{4}$

微积分

$\ \lim _{x\to -1}{\frac {1-x}{1+x}}$

$\ \lim _{x\to 1}{\frac {x^{2}+1}{{\sqrt {2+2x}}-2}}=4$

$\ {\frac {x^{2}+1}{{\sqrt {2+2x}}-2}}{\frac {{\sqrt {2+2x}}+2}{{\sqrt {2+2x}}+2}}={\frac {(x^{2}+1)({\sqrt {2+2x}}+2)}{2x-2}}$

$\ {\sqrt {2}}$
$\ {\sqrt {2+{\sqrt {2}}}}$
$\ {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}$
$\ {\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}$

物理

$\ mc^{2}-m_{0}c^{2}=(\gamma -1)m_{0}c^{2}$

$\ u={\frac {v_{1}+v_{2}}{1+{\frac {v_{1}v_{2}}{c^{2}}}}}$

$\ U(r)=-r_{0}U_{0}({\frac {e^{-{\frac {r}{r_{0}}}}}{r}}),F_{r}=-{\frac {dU(r)}{dr}}=-({\frac {U_{0}}{r}})(1+{\frac {r_{0}}{r}})e^{-{\frac {r}{r_{0}}}}$

$\ (a.bc\pm d)\times 10$

$\ (m_{1}+m_{2})v_{1}^{2}+(m_{1}-m_{2})u_{1}^{2}-(2m_{1}u_{1}cos\theta _{1})v_{1}=0,cos^{2}\theta _{1}=1-({\frac {m_{2}}{m_{1}}})^{2}$

$\ x_{CM}={\frac {\int x(\,\lambda dx)}{\int \,\lambda dx}}$

$\ v=-\omega Rsin(\omega t)$

$\ v_{f}=1+2ln{\frac {50}{47}}=1.12km/s$

$\ v_{f}-v_{i}=({\frac {F}{m}})t=\int _{0}^{30}{\frac {2\times 10^{5}}{50000-100t}}\,dt=\int _{0}^{30}{\frac {2000}{500-t}}\,dt=123.75m/s$

$\ {\vec {F}}=-k{\vec {v}}^{2}$

$\ \lim _{x\to 0}xsin({\frac {1}{x}})=0$

${\frac {1}{\lambda }}=R({\frac {1}{n^{2}}}-{\frac {1}{n'^{2}}})$

$\ f={\frac {mv}{4\ell }}$

$\ \int _{0}^{sue}f(x)\,dx$

$\ \alpha ={\frac {m}{2kT}}={\frac {M}{2RT}}={\frac {28\times 10^{-3}}{2(8.314)(300)}}=5.6\times 10^{-6}$

$\ \ell =m{\frac {\lambda }{4}}$

$\ v=f\lambda$

物理实验

$\ y\pm {\frac {\sigma }{y}}\times 100\%$

$\ E={\frac {1}{2}}\mu (u-0)^{2},{\frac {1}{\mu }}={\frac {1}{m}}+{\frac {1}{M}}$

$\ a={\frac {v^{2}}{r}}$

$\ F={\frac {mv^{2}}{r}}=mr\omega ^{2}$

$\ v={\frac {2\pi r}{T}}=r\omega$

$\ F={\frac {4\pi ^{2}r}{T^{2}}}m$ , $\ F\propto m$

$\ r=({\frac {F}{4\pi ^{2}m}})T^{2}$ , $\ T\propto {\sqrt {r}}$

$\ F=(4\pi ^{2}mr){\frac {1}{T^{2}}}$ , $\ F=(\not 4\not \pi ^{2}\not m\not r){\frac {1}{T^{2}}}$ , $\ F\propto {\frac {1}{T^{2}}}$

$\ {\vec {F}}=m{\vec {a}}$

$\ M_{2}g=(M_{1}+M_{2}+{\frac {m}{2}})a,a={\frac {M_{2}g}{(M_{1}+M_{2}+{\frac {m}{2}})}}$

$\ a={\frac {{\frac {5}{\Delta t_{1}}}-{\frac {5}{\Delta t_{2}}}}{\Delta t_{1}-\Delta t_{2}}}={\frac {5}{(\Delta t_{1}-\Delta t_{2})^{2}}}$

$\ a={\frac {S_{1}-S_{2}}{T^{2}}}=(S_{1}-S_{2})f^{2}$

$\ {\begin{cases}M_{2}a=M_{2}g-T_{2}\\M_{1}a=T_{1}\end{cases}}$

$\ \tau =I\alpha =({\frac {1}{2}}mr^{2})({\frac {a}{r}})={\frac {1}{2}}mra$

$\ r(T_{2}-T_{1})=\tau$

$\ \not {r}(M_{2}g-M_{2}a-M_{1}a)={\frac {1}{2}}m\not {r}a$

$\ I={\frac {r^{2}T}{a}}={\frac {mr^{2}(g-a)}{a}}$

$\ mg-T=ma$

$\ v={\frac {F}{\mu }}$

正切半角

设 $\ tan{\frac {A}{2}}$ = t
$\ sin2A=2sinAcosA=2{\frac {2t}{1+t^{2}}}{\frac {1-t^{2}}{1+t^{2}}}={\frac {1}{4}}t$

$\ \int e^{-x}cschx\,dx$
$\ \int xe^{-x}cschx\,dx$

重量

$\ {\frac {w-85}{6.5-6}}={\frac {90-85}{6.3-6}},w=93.33$

机制

$\ M={\frac {PL}{4}}$

$\ I={\frac {\pi d^{4}}{64}}$

MOSFET

$\ {\frac {\partial {\sqrt {Id}}}{\partial Vgs}}={\frac {\partial (Vgs-Vth){\sqrt {K_{n}(1+\lambda Vds)}}}{\partial Vgs}}$

$\ {\sqrt {Id}}=(Vgs-Vth){\sqrt {K_{n}(1+\lambda Vds)}}$

$\ g_{m}={\frac {2Id}{Vgs-Vth}}$

feedback control 4.16 d

$\ E_{w}=-V=-{\frac {\frac {m}{s}}{1+{\frac {k_{3}m}{s}}+{\frac {k_{1}k_{2}m}{s^{2}}}}}=-{\frac {ms}{s^{2}+k_{3}ms+k_{1}k_{2}m}}$

$\ =-{\frac {1}{1+{\frac {s^{2}+k_{3}ms+k_{1}k_{2}m-ms}{ms}}}}=-{\frac {1}{1+G}}$

$\ G={\frac {s^{2}+k_{3}ms+k_{1}k_{2}m-ms}{ms}}$

$\ k_{p}=\lim _{s\to 0}G=\lim _{s\to 0}{\frac {s^{2}+k_{3}ms+k_{1}k_{2}m-ms}{ms}}=\infty$

$\ k_{v}=\lim _{s\to 0}sG=\lim _{s\to 0}{\frac {s^{2}+k_{3}ms+k_{1}k_{2}m-ms}{m}}=k_{1}k_{2}$

Ad

$\ A_{d}={\frac {V_{0}}{V_{id}}}=(g_{m}R_{0})({\frac {1}{1+sR_{0}C_{L}}})[{\frac {1+s({\frac {C_{m}}{2g_{m3}}})}{1+s({\frac {C_{m}}{g_{m3}}})}}]$

反甲

$\ z\times {\frac {100}{(100+x)}}=z\times 0.3\times {\frac {100}{(100+y)}}$

$\ {\frac {1}{(100+x)}}={\frac {0.3}{(100+y)}}$

$\ {\frac {w}{(100+x)}}={\frac {0.3}{(100+y)}}$

sys

$\ sys1={\frac {1}{(4s^{2}+18s+9)}},\zeta =1.5$

$\ sys2={\frac {1}{(4s^{2}+12s+9)}},\zeta =1$

$\ sys3={\frac {1}{(4s^{2}+6s+9)}},\zeta =0.5$

$\ sys4={\frac {1}{(4s^{2}+3s+9)}},\zeta =0.25$