Question #ffc58

2 Answers
Jim G.
Aug 27, 2017

#"see explanation"#

Explanation:

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

#•color(white)(x)cos(A+B)=cosAcosB-sinAsinB#

#•color(white)(x)sin(A-B)=sinAcosB-cosAsinB#

#•color(white)(x)cos2A=cos^2A-sin^2A#

#"consider left hand side"#

#sinx(cosxcosy-sinxsiny)-cosx(sinxcosy-cosxsiny)#

#=cancel(sinxcosxcosy)-sin^2xsinycancel(-sinxcosxcosy)+cos^2xsiny#

#=siny(cos^2x-sin^2x)#

#=sinycos2x=" right hand side "rArr" proved"#

Tazwar Sikder
Aug 27, 2017

We have: #sin(x) cos(x + y) - cos(x) sin(x - y)#

Let's apply the compound angle identities for #sin(x)# and #cos(x)#:

#= sin(x) cdot (cos(x) cos(y) + sin(x) sin(y)) - cos(x) cdot (sin(x) cos(y) - cos(x) sin(y))#

#= sin(x) cos(x) cos(y) + sin^(2)(x) sin(y) - sin(x) cos(x) cos(y) - cos^(2)(x) sin(y)#

#= sin^(2)(x) sin(y) - cos^(2)(x) sin(y)#

#= sin(y) (sin^(2)(x) - cos^(2)(x))#

#= sin(y) (- (cos^(2)(x) - sin^(2)(x)))#

#= - sin(y) (cos^(2)(x) - sin^(2)(x))#

Then, let's apply the double angle identity for #cos(x)#; #cos(2 x) = cos^(2)(x) - sin^(2)(x)#:

#= - sin(y) cdot cos(2 x)#

#= - cos(2 x) sin(y)#

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