Question 76

If $$\sin A = \frac{1}{2}$$ and $$A$$ and $$B$$ are complementary angles, then find the value of $$\sin (A - B)$$.

Solution

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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