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

2 Answers
Lucy
May 26, 2018

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

Jim G.
May 26, 2018

#"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"#

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