MCQ
$\cos 405^{\circ}=?$
  • A
    $\frac{-1}{\sqrt{2}}$
  • B
    $-\sqrt{2}$
  • C
    $\sqrt{2}$
  • D
    $\frac{1}{\sqrt{2}}$

Answer

(d)$\frac{1}{\sqrt{2}}$
Explanation: $\cos 405^{\circ}=\cos \left(360^{\circ}+45^{\circ}\right)=\cos 45^{\circ}=\frac{1}{\sqrt{2}}$

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