How will you prove the formula #sin3A=3sinA-4sin^3A# using only the identity #sin(A+B)=sinAcosB+cosAsinB#?

Question

31723 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-23T03:36:10

The Given Identity

#color(blue)(sin(A+B)=sinAcosB+cosAsinB........(1)#

#"Inserting " color(red)(-B) " in place of B" #

#color(red)(sin(A+B)=sinAcosB+cosAsinB........(2)#

Adding (2) to (1)

#color(green)(sin(A+B)+sin(A-B)=2sinAcosB.......(3)#

Putting 2A in place of B in (3)

#color(blue)(sin(A+2A)+sin(A-2A)=2sinAcos2A#

#=>color(green)(sin3A-sinA=2sinAcos2A#

#=>color(green)(sin3A=2sinAcos2A+sinA................(4)#

Putting (90-A) in place of B in (3)

#color(blue)(sin(A+90-A)+sin(A-90+A)=2sinAcos(90-A)#

#=>color(blue)(sin(90)-sin(90-2A)=2sinAsinA#

#=>color(blue)(1-cos2A=2sin^2A)#

#=>color(blue)(cos2A=1-2sin^2A)#

#"Putting "color(red)(cos2A=1-2sin^2A) " in "(4)#

#=>color(green)(sin3A=2sinAcos2A+sinA#

#=>color(green)(sin3A=2sinA(1-2sin^2A)+sinA.#

#:.color(red)(sin3A=3sinA-4sin^3A).#

Proved