Prove cos3x=4cos3x−3cosx
Proof.
Consider,
L.H.S=cos3x=cos(2x+x)∵cos(A+B)=cosAcosB−sinAsinB=cos2xcosx−sin2xsinx∵cos2x=2cos2x-1andSin2x=2sinxcosx=(2cos2x-1)cosx−2cosxsinxsinx=2cos3x-cosx−2sin2xcosx=-cosx+2cos3x−2sin2xcosx=-cosx+2cos3x−2sin2xcosx∵sin2x=1-cos2x=-cosx+2cos3x−2(1-cos2x)cosx=-cosx+2cos3x−2cosx+2cos3x=4cos3x−3cosx=R.H.S
Hence, proved.
prove that: