Some Help Please - Thanks in advance

Question

Hello,

There are a few questions that I am not sure how to do:

1) sin2x cos2x - 2cos2x sinx = sinx

2) [sin(A+B)] / [sin(A-B)] = [tanA + tanB] / [tanA - tanB]

3) tanA + tanB = [sin(A+B)] / [cosA cosB]

Thank you so much.

Answer 1

2)

[sin(A+B)] / [sin(A-B)] = [tanA + tanB] / [tanA - tanB]

Take left hand identity:

[sin(A+B)] / [sin(A-B)]=

Using angle sum and difference identities:

sin(A+B)=sinAcosB+cosAsinB

sin(A-B)=sinAcosB-cosAsinB

[sin(A+B)] / [sin(A-B)]=(sinAcosB+cosAsinB)/(sinAcosB-cosAsinB)

Divide numerator and denominator by cos A cosB.

=(sinAcosB+cosAsinB)/(cos A cosB)/(sinAcosB-cosAsinB)/(cos A cosB)

=((sinAcosB)/(cos A cosB)+(cosAsinB)/(cos A cosB))/((sinAcosB)/(cos A cosB)-(cosAsinB)/(cos A cosB))

=((sinA/cosA)+(sinB/cosB))/((sinA/cosA)-(sinB/cosB))

=(tanA+tanB)/((tanA-tanB)

---------------------------

3)

tanA + tanB = [sin(A+B)] / [cosA cosB]

Take left hand identity:

tanA + tanB =(sinA/cosA)+(sinB/cosB)

                           = (sinAcosB+cosAsinB)/(cosAcosB)     

Using angle sum and difference identities:

sin(A+B)=sinAcosB+cosAsinB

=sin(A+B)/(cosAcosB)

Answer 2

1) sin2x cosx - cos2x sinx = sinx

Take left hand identity:

sin2x cosx - cos2x sinx

Using angle difference identity:

sin(A-B)=sinAcosB-cosAsinB

Here A=2x and B=x

sin2x cosx - cos2x sinx= sin(2x - x)

                                      = sinx