Question

How to prove that `sin(A+B)sin(A-B)=sin^2A-sin^2B`?

Answer 1

Below

Explanation:

`sin(A+B)sin(A-B)=sin^2A-sin^2B`

LHS

= `sin(A+B)sin(A-B)`

Recall: `sin(alpha-beta)=sinalphacosbeta-cosalphasinbeta`
And `sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta`

= `(sinAcosB+cosAsinB)times(sinAcosB-cosAsinB)`

= `sin^2Acos^2B-cos^2Asin^2B`

Recall: `sin^2alpha+cos^2alpha=1`
From above, we can then assume correctly that :

`sin^2alpha=1-cos^2alpha` AND
`cos^2alpha=1-sin^2alpha`

= `sin^2A(1-sin^2B)-sin^2B(1-sin^2A)`

= `sin^2A-sin^2Asin^2B-sin^2B+sin^2Asin^2B`

= `sin^2A-sin^2B`

= RHS

Answer 2

`"see explanation"`

Explanation:

`"using the "color(blue)"trigonometric identities"`

`•color(white)(x)sin(x+y)=sinxcosy+cosxsiny`

`•color(white)(x)sin(x-y)=sinxcosy-cosxsiny`

`"consider the left side"`

`(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)`

`=sin^2Acos^2B-cos^2Asin^2B`

`=sin^2A(1-sin^2B)-sin^2B(1-sin^2A)`

`=sin^2Acancel(-sin^2Asin^2B)-sin^2Bcancel(+sin^2Asin^2B)`

`=sin^2A-sin^2B`

`="right side "rArr"verified"`