Answer
$0$
Work Step by Step
Need to solve $\cos[ (2n+1) \times (90^{\circ}) ] $
In order to solve the given expression we can use Trigonometric table of angles.
We know that $\sin (-\theta) =- \sin \theta$ and $\cos (-\theta) =\cos \theta$
Set $n=0, 1,2,3, .....$
Now, $\cos[ (2(0)+1) \times (90^{\circ}) ] =\cos (90^{\circ}) =0$
$\cos[ (2(1)+1) \times (90^{\circ}) ] =\cos[ (3) \times (90^{\circ}) ] =\cos 270^{\circ} =0$
$\cos[ (2(3)+1) \times (90^{\circ}) ] =\cos[ (6) \times (90^{\circ}) ] =\cos 540^{\circ} = 0$
Therefore, our answer is: $\cos[ (2n+1) \times (90^{\circ}) ] =0$
