Question

sinx-sin3x/sin^2x-cos^2x=?​

Answer 1

Given Expression,

$\sf{ \frac{sin \: x - sin3x}{sin {}^{2}x - cos {}^{2}x } }$

Now,

Numerator:

sin x - sin3x

Of the form,

sin X - sin Y = 2cos(X+Y)/2sin(X-Y)/2

=2cos2xsin(-x)

= - 2cos2xsin x

Denominator:

sin²x-cos²x

= -(cos²x-sin²x)

= -cos2x

Implies,

$\sf{ \frac{ - 2cos2x.sin \: x}{ - cos2x} } \\ \\ = \sf{sin \: x}$