trigonometry formulae

can someone explain to me please how you can use sin(A+/-B) with A=3x and B=x to show 2sinxcos3x can be written as sinpx-sinqx

Reply 1

Gaz031

Freeway

can someone explain to me please how you can use sin(A+/-B) with A=3x and B=x to show 2sinxcos3x can be written as sinpx-sinqx

sin(A+B)-sin(A-B)=[sinAcosB+cosAsinB]-[sinAcosB-cosAsinB]
=2cosAsinB
If A=3x, B=x.
2cos3xsinx=sin4x-sin2x.

Reply 2

Christophicus

Freeway

can someone explain to me please how you can use sin(A+/-B) with A=3x and B=x to show 2sinxcos3x can be written as sinpx-sinqx

sin(A+B) = sinAcosB + sinBcosA
sin(A-B) = sinAcosB - sinBcosA

sin(A+B) + sin(A-B) = 2sinAcosB
2sinxcos3x = sin(x+3x) + sin(x-3x) = sin(4x) + sin(-2x)

sin(-@) = -sin@
2sinxcos3x = sin(4x) - sin(2x)

Reply 3

Freeway

cheers, i just couldnt figure it out, even though now i see it, it is quite obvious, thanks a lot!!

Reply 4

matt.forster

2cosAsinB is the same as 2sinBcosA
so... 2cosAsinB≡sin(A+B)- sin(A-B)
A=3x B=x
2sinxcos3x= sin(3x+x) - sin(3x-x)
=sin4x-sin2x

think this is right

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