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Mathematics Question on Trigonometric Functions of Sum and Difference of Two Angles

Prove that cos9x−cos5xsin 17x−sin3x=−sin 2xcos 10x\frac{cos 9x-cos5x}{sin\,17x-sin3x}=-\frac{sin\,2x}{cos\,10x}

Answer

It is known that

cosA−cosB=−2sin(A+B2)sin(A−B2),sinA−sinB=2cos(A+B2)sin(A−B2)cosA-cosB=-2sin(\frac{A+B}{2})sin(\frac{A-B}{2}),sinA-sinB=2cos(\frac{A+B}{2})sin(\frac{A-B}{2})

L.H.S=cos9x−cos5xsin17x−sin3xL.H.S=\frac{cos9x-cos5x}{sin17x-sin3x}

=−2 sin(9x+5x2).sin(9x−5x2)2cos(17x+3x2).sin(17x−3x2)=\frac{-2\,sin(\frac{9x+5x}{2}).sin(\frac{9x-5x}{2})}{2cos(\frac{17x+3x}{2}).sin(\frac{17x-3x}{2})}

=−2 sin7x.sin2x2cos10x.sin7x=\frac{-2\,sin7x.sin2x}{2cos10x.sin7x}

=−sin2xcos10x=-\frac{sin2x}{cos10x}

=R.H.S=R.H.S