fisica

TALLER DE FÍSICA 1

CUAL DEBE SER EL REMPLAZO DE.

sen[90-b]=sen90.cosb-cos90.senb=1.cosb-0.senb=cosb-0= = cosb RTA.sen[90+b]=sen90.cosb+cos90.senb=1.cosb+0senb=cosb+0= =cosb RTA

cos[90-b]=cos90.cosb+sen90.senb=0.cosb+1.senb=0+senb= =senb RTAtg[90+b]=(tg90+tgb)/(1-(tg90.tgb) )=(sen90/( cos90 )+)/(sen90/cos90) (senb/cosb)/(+senb/1cosb)=((sen90.cosb)+(senb.cos90))/(cos90.cosb)=(-1)/1 ( sen90.cosb+senb)/(cos90.cosb)=(sen90.cosb+senb.cos90)/(cos90.cosb)=( cos90.cosb-sen90.senb)/(cos90.cosb)=( (sen90.cos90)+(senb.cos90)/((cosb.cos90)-(sen90.senb))=( cosb+0)/(0-1senb)=( cosb)/(-senb)=-cotgb RTA

VERIFICARtgγ+secγ=(secγ-1+tgγ)/(1-secγ+tgγ)=no se cumple porque.
(secγ-1+tgγ)/(1-secγ+tgγ)=(1/( cosγ )-)/(1/) (1/)/(-1/cosγ) (+senγ/cosγ)/(+senγ/( cosγ ))=((1-cosγ+senγ)/( cosγ))/((cosγ-1+senγ)/cosγ)=(1-cosγ+senγ)/(cosγ-1+senγ) rta
tgγ+secγ=senγ/cosγ+1/cscγ=(1/( cscγ )-)/(1/secγ) (1/)/(+1/senγ)=secγ/(cscγ+secγ) rta
DEMUESTRE
(secθ-cscθ)/(cscθ+secθ)=(1+tgθ-2)/(tgθ+1)=nosecumple porque

(secθ-cscθ)/(cscθ+secθ)=(1/( cosθ )-)/(1/senθ) (1/senθ)/(+1/( cosθ ))=(senθ/( cosθ )-)/(cosθ1/senθ) (( cosθ)/senθ)/(+senθ/( cosθ ))= (senθ-cosθ)/(cosθ+senθ)rta


(1+tgθ-2)/(tgθ+1)=(1/( 1 )+)/(senθ/cosθ) (senθ/cosθ)/(+1/( 1 ))=((cosθ+senθ-2cosθ)/( cosθ))/((senθ+cosθ)/cosθ)=(cosθ+senθ-2cosθ)/(senθ+cosθ)=(senθ-cosθ)/(senθ+cosθ)rta




sen[180-b]=sen180.cosb-cos180.senb=0.cosb-(-1).senb=0-(-senb)=senb RTA


sen[180+b]=sen180.cosb+cos180.senb=0.cosb+(-1).senb=0.cosb+(-senb)=-senb RTAcos[180-b]=cos180.cosb+sen180.senb=(-1).cosb+0= =-cosb RTA

tg[90-b]=(tg90-tgb)/(1+tg90.tgb)=(sen90/(cos90 )-)/(1+(sen90/cos90)...