If sin (A - B) = \(\frac{1}{2}\) and cos (A + B) = \(\frac{1}{2}\) , what is the value of sin A. sin 2B + sin2 A . sin B cos B?
Answer & explanation
Correct answer: option 1
sin (A - B) = \(\frac{1}{2}\) ⇒ sin (A - B) = sin 30°
A - B = 30° ...........(i)
cos (A + B) = \(\frac{1}{2}\) ⇒ cos (A + B) = cos 60°
A + B = 60° ............(ii)
Adding (i) , (ii)
2A = 90° ⇒ A = 45°
and B = 15°
sin A. sin 2B + sin2 A. sin B cos B
= sin 45° sin 30° + (sin2 45°) × \(\frac{1}{2}\) × (2sin 15°.cos 15°)
= \(\frac{1}{\sqrt {2 }}\) × \(\frac{1}{2}\) + \(\frac{1}{2}\) × \(\frac{1}{(\sqrt {2 }})^2\) (sin 30°)
= \(\frac{1}{2\sqrt { 2}}\) + \(\frac{1}{4}\) × \(\frac{1}{2}\) = \(\frac{4 + \sqrt {2 }}{8\sqrt {2 }}\)