MCQ
The value of $\cos52^\circ+\cos68^\circ+\cos172^\circ\text{ is }$
  • $0$
  • B
    $1$
  • C
    $2$
  • D
    $\frac{3}{2}$

Answer

Correct option: A.
$0$
$\cos52^\circ+\cos68^\circ+\cos172^\circ$
$=\ 2\cos\Big(\frac{52^\circ+68^\circ}{2}\Big)\cos\Big(\frac{52^\circ-68^\circ}{2}\Big)+\cos172^\circ$ $\Big[\because\ \cos\text{A}+\cos\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=\ 2\cos60^\circ\cos(-8^\circ)+\cos172^\circ$
$=\ 2\times\frac{1}{2}\cos8^\circ+\cos172^\circ$
$=\ \cos8^\circ+\cos172^\circ$
$=\ 2\cos\Big(\frac{8^\circ+172^\circ}{2}\Big)\cos\Big(\frac{8^\circ-172^\circ}{2}\Big)$
$=\ 2\cos90^\circ\cos82^\circ$
$=\ 0$

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