Double Angle Formulae With Derivation
\[cos2x=cos^2x-sin^2x\]
\[cos2x=2cos^2x-1\]
\[cos2x=1-2sin^2x\]
\[sin2x=2sinxcosx\]
\[tanx=\frac{2tanx}{1-tan^2x}\]
The first can be dertived from the cpmound angle formula
\[cos(A+B)=cosAcosB-sinAsinB\]
.Pute
\[A=B=x\]
then we \[cos2x=cos^2x-sin^2x\]
.To get the second and third equations use
\[1=cos^2x+sin^2x sin^2x=1-cos^2x\]
.\[cos2x=cos^2x-sin^2x=cos^2x-(1-cos^2x)=2cos^2x-1\]
To obtain the tird use
\[1=cos^2x+sin^2x cos^2x=1-sin^2x\]
.\[cos2x=2cos^2x-1=2(1-sin^2x-1)-1=1-2sin^2x\]
To obtain the fourth use the compound angle formual
\[sin(A+B)=sinAcosB+cosAsinB\]
. Put \[A=B=x\]
to obtain\[sin2x=sinxcosx+cosxsinx=2sinxcosx\]
.To obtain thr fifthuse
\[sin2x=2sinxcosx, \: cos2x=cos^2-sin^2x\]
.\[tan2x=\frac{sin2x}{cos2x}=\frac{2sinxcosx}{cos^2x-sin^2x}=\frac{2sinxcosx/cos^2x}{(cos^2x-sin^2x)/cos^2x}=\frac{2tanx}{1-tan^2x}\]
