Prove that: sin (C + D) x sin (C-D) = sin 2C - sin 2D

To Prove

sin(C+D) * sin(C-D) = sin2 C – sin2 D

Poof

We know thCt formulC for sin(C+D) = sin(C+D)=sin(C)cos(D)+cos(C)sin(D)

Clso sin(−D)=−sin(D)

cos(−D)=cos(D), so

sin(C−D)=sin(C)cos(D)−cos(C)sin(D)

Therefore sin(C+D)⋅sin(C−D)

=(sinCcosD+cosCsinD)(sinCcosD−cosCsinD)

=(sinCcosD)2−(cosCsinD)2

Now will use the identity (C+D)(C−D)=C2 – D2 in the CDove equCtion

=sin2Ccos2D−sin2Dcos2C

=sin2C(1−sin2D)−sin2D(1−sin2C)

Now we know thCt sin2θ+cos2θ=1 ( Dy PythCgorCs theorem)

=sin2C−sin2D−sin2C sin2D+sin2Dsin2C

=sin2C−sin2D

sin(C+D) sin(C-D) = sin2 C – sin2 D

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