How can verify this equation is an identity? Sin(α+ß)Sin(α-ß)=Sin²α-Sin²ß

3 Answers
Noah G · Ratnaker Mehta
Mar 20, 2018

See below.

Explanation:

Use the addition and subtraction formulas for sine.

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

Now

#(sin alphacosbeta+ sinbetacosalpha)(sinalphacosbeta - sinbetacosalpha) = sin^2alpha - sin^2beta#

On the left it's clearly a difference of squares.

#sin^2alphacos^2beta - sin^2betacos^2alpha = sin^2alpha - sin^2beta#

#sin^2alpha(1 - sin^2beta) - sin^2beta(1 - sin^2alpha) = sin^2alpha - sin^2beta#

#sin^2alpha - sin^2alphasin^2beta - sin^2beta + sin^2alphasin^2beta = sin^2alpha - sin^2beta#

#sin^2alpha - sin^2beta = sin^2alpha - sin^2beta#

As required.

Hopefully this helps!

maganbhai P.
Mar 20, 2018

The proof of the identity is given below.

Explanation:

We know that,

#color(red)((1)sin(x+y)=sinxcosy+cosxsiny#

#color(red)((2)sin(x-y)=sinxcosy-cosxsiny#

Using #color(red)((1) and (2)# ,we get

#LHS=sin(alpha+beta)sin(alpha-beta)#

#=(sinalphacosbeta+cosalphasinbeta)(sinalphacosbeta-cosalphasinbeta)#

#=sin^2alphacolor(red)(cos^2beta)-color(blue)(cos^2alpha)sin^2beta#

#=sin^2alphacolor(red)((1-sin^2beta))-color(blue)((1-sin^2alpha))sin^2beta#

#=sin^2alpha-cancel(sin^2alphasin^2beta)-sin^2beta+cancel(sin^2alphasin^2beta)#

#=sin^2alpha-sin^2beta#

#=RHS#

Ratnaker Mehta
Mar 20, 2018

Kindly go through a Proof in the Explanation.

Explanation:

The following is another way to prove the assertion :

Recall that, #-2sinxsiny=cos(x+y)-cos(x-y)#.

#:. sin(alpha+beta)sin(alpha-beta)#,

#=-1/2{-2sin(alpha+beta)sin(alpha-beta)}#,

#=-1/2{cos((alpha+beta)+(alpha-beta))-cos((alpha+beta)-(alpha-beta))}#,

#=-1/2{cos2alpha-cos2beta}#,

#=-1/2{(1-2sin^2alpha)-(1-2sin^2beta)}#,

#=sin^2alpha-sin^2beta#.

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