Question

${\sum }_{r=1}^{16}\sqrt{1+cos4r}=$

• A. $4\sqrt{2}sin{1}^{\circ }cos{2}^{\circ }cos{17}^{\circ }$
• B. $4\sqrt{2}cos{1}^{\circ }cos{2}^{\circ }cos{17}^{\circ }$
• C. $4\sqrt{2}cos{1}^{\circ }sin{2}^{\circ }cos{17}^{\circ }$
• D. None of these

Answer 1

The correct option is D

None of these

${\sum }_{r=1}^{16}\sqrt{1+cos4r}={\sum }_{r=1}^{16}\sqrt{2}cos2r$

$=\sqrt{2}(cos{2}^{\circ }+cos{4}^{\circ }+cos{6}^{\circ }+\cdots \cdots +cos{32}^{\circ })$

$=\sqrt{2}\frac{sin{16}^{\circ }}{sin{1}^{\circ }}cos\{2+(15\left)\right(1\left)\right\}=\sqrt{2}\frac{sin{16}^{\circ }cos{17}^{\circ }}{sin{1}^{\circ }}$