The Student Room Group
Reply 1
sin2x+cos2x=1cosecx=1sinxsecx=1cosxcotx=1tanx1+tan2x=sec2x1+cot2x=cosec2xsin(A+B)=sinAcosB+cosAsinBcos(A+B)=cosAcosBsinAsinBtan(A+B)=tanA+tanB1tanAtanB(You dont really need to remember the (AB) or double/half angle formulae)2sinAcosB=sin(A+B)+sin(AB)\LARGE sin^2x+cos^2x=1\\cosecx=\frac{1}{sinx}\\secx=\frac{1}{cosx}\\cotx=\frac{1}{tanx}\\1+tan^2x=sec^2x\\1+cot^2x=cosec^2x\\sin(A+B)=sinAcosB+cosAsinB\\cos(A+B)=cosAcosB-sinAsinB\\tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}\\(You \ don't \ really \ need \ to \ remember \ the \ (A-B) \ or \ double/half \ angle \ formulae)\\2sinAcosB=sin(A+B)+sin(A-B)
2cosAsinB=sin(A+B)sin(AB)2cosAcosB=cos(A+B)cos(AB)2sinAsinB=cos(AB)cos(A+B)sinP+sinQ=2sinP+Q2cosPQ2sinPsinQ=2cosP+Q2sinPQ2cosP+cosQ=2cosP+Q2cosPQ2cosQcosP=2sinP+Q2sinPQ2\LARGE 2cosAsinB=sin(A+B)-sin(A-B)\\2cosAcosB=cos(A+B)-cos(A-B)\\2sinAsinB=cos(A-B)-cos(A+B)\\sinP+sinQ=2sin\frac{P+Q}{2}cos\frac{P-Q}{2}\\sinP-sinQ=2cos\frac{P+Q}{2}sin\frac{P-Q}{2}\\cosP+cosQ=2cos\frac{P+Q}{2}cos\frac{P-Q}{2}\\cosQ-cosP=2sin\frac{P+Q}{2}sin\frac{P-Q}{2}
That's the bulk of P2 trig
Reply 2
hey, cheers mate!