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Sagot :
The question says we should prove that:
sin(x+y)cosy-cos(x+y)siny-sinx = 0
The only way we can prove this is using trigonotric identity.
recall:
sin(x + y) = sinxcosy + cosxsiny
cos(x + y) = cosxcosy + sinxsiny
So,
(sinxcosy + cosxsiny)cosy - (cosxcosy - sinxsiny)siny-sinx = 0
Opening the brackets
sinxcos^2y + cosxsinycosy - (cosxcosysiny - sinxsin^2y) - sinx = 0
sinxcos^2y + cosxsinycosy - cosxcosysiny + sinxsin^2y - sinx = 0
some identity cancels out.
sinxcos^2y + sinxsin^2y - sinx = 0
factoring out.
sinx (cos^2y + sin^2y) - sinx = 0
let cos^2y + sin^2y = 1
sinx (1) - sinx = 0
collecting like terms
sinx - sinx = 0
the sinx cancels out
0 = 0
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