Question

Without using tables, show that (cos ${35}^{\circ }$ cos ${55}^{\circ }$ - sin ${35}^{\circ }$ sin ${55}^{\circ }$) = 0

Answer 1

LHS = (cos ${35}^{\circ }$ cos ${55}^{\circ }$ - sin ${35}^{\circ }$ sin ${55}^{\circ }$)

= [(cos ${35}^{\circ }$ cos ${55}^{\circ }$ - sin( ${90}^{\circ }$ - ${55}^{\circ }$) sin( ${90}^{\circ }$ - ${35}^{\circ }$)]

= (cos ${35}^{\circ }$ cos ${55}^{\circ }$ - cos ${55}^{\circ }$ cos ${35}^{\circ }$ ) = 0 = RHS [$\because sin(90-\theta )=cos\theta andcos(90-\theta )=sin\theta$ ]