MCQ
$\cos40^\circ+\cos80^\circ+\cos160^\circ+\cos240=$
- A$0$
- B$1$
- C$\frac{1}{2}$
- D$-\frac{1}{2}$
Solution:$$
$\cos40^\circ+\cos80^\circ+\cos160^\circ+\cos240^\circ$
$=\ 2\cos\Big(\frac{40^\circ+80^\circ}{2}\Big)\cos\Big(\frac{40^\circ-80^\circ}{2}\Big)\\ \ \ \ +\cos160^\circ-\cos(180^\circ+60^\circ)$$\big[\because\ \cos\text{A}+\cos\text{B}=2\big]$
$=\ 2\cos60^\circ\cos(-20^\circ)+\cos160^\circ-\frac{1}{2}$
$=\ 2\times\frac{1}{2}\cos20^\circ+\cos160^\circ-\frac{1}{2}$
$=\ -\cos(180-20)^\circ+\cos160^\circ-\frac{1}{2}$
$=\ -\frac{1}{2}$
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A die is rolled, then the probability that an even number is obtained is: