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Prove that sin(x+y)cosy-cos(x+y)siny-sinx=0

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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