If $$\sin A = \frac{1}{2}$$ and $$A$$ and $$B$$ are complementary angles, then find the value of $$\sin (A - B)$$.
According to question, $$A+B=90^{\circ\ }.$$
$$\sin\left(A-B\right)=\sin A\cos B-\sin B\cos A$$
=$$SinASinA-\left(CosA\right)CosA.\ \ \left(As\ Cos\left(90-A\right)=SinA\right).$$
=$$Sin^2A-Cos^2A.$$
=$$Sin^2A-\left(1-Sin^2A\right).$$
=$$2Sin^2A-1.$$
=$$2\left(\frac{1}{2}\right)^2-1.$$
=$$\frac{1}{2}-1.$$
=$$-\frac{1}{2}.$$
C is correct choice.
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