Question

$\sin {18}^{\circ }\cdot \cos {39}^{\circ }+\sin 6^{\circ }\cdot \cos {15}^{\circ }=-\sin {24}^{\circ }\cdot \cos {33}^{\circ }$.

• A. True
• B. False

Answer 1

The correct option is B False

$LHS=\sin {18}^{\circ }\cos {39}^{\circ }+\sin 6^{\circ }\cos {15}^{\circ }$

$=\frac{1}{2}[2\sin {18}^{\circ }\cos {39}^{\circ }+2\sin 6^{\circ }\cos {15}^{\circ }]$

We know that $2\sin A\cos B=\sin (A+B)+\sin (A-B)$

$=\frac{1}{2}[\sin ({18}^{\circ }+{39}^{\circ })+\sin ({18}^{\circ }-{39}^{\circ })+\sin (6^{\circ }+{15}^{\circ })+\sin (6^{\circ }-{15}^{\circ })]$

$=\frac{1}{2}[\sin {47}^{\circ }-\sin {21}^{\circ }+\sin {21}^{\circ }-\sin 9^{\circ }]$

$=\frac{1}{2}[\sin {47}^{\circ }-\sin 9^{\circ }]$

$=\frac{1}{2}\times 2\cos \left(\frac{47+9}{2}\right)\sin \left(\frac{47-9}{2}\right)$

$=\cos \left(\frac{56}{2}\right)\sin \left(\frac{38}{2}\right)$

$=\cos {28}^{\circ }\sin {19}^{\circ }$

Hence $LHS\ne RHS$

Hence the given statement is false.