Question

The value of the $\sin 1^0+\sin 2^0+.....+\sin {359}^0$ is equal to

• A. $0$
• B. $1$
• C. $-1$
• D. $180$

Answer 1

The correct option is A $0$

The trignometric function is $\sin 1^0+\sin 2^0+\sin 3^0+.....+\sin {358}^0+\sin {359}^0$.

Rewrite the above function as:

$\sin 1^0+\sin 2^0+\sin 3^0+.....+\sin (360-3{)}^0+\sin (360-2{)}^0+\sin (360-1{)}^0$

Apply periodic property $\sin (360-\theta {)}^0=-\sin \theta$ as shown below:

$=\sin 1^0+\sin 2^0+\sin 3^0+.....+(-\sin 3^0)+(-\sin 2^0)+(-\sin 1^0)=\sin 1^0+\sin 2^0+\sin 3^0+.....-\sin 3^0-\sin 2^0-\sin 1^0=\sin 1^0-\sin 1^0+\sin 2^0-\sin 2^0+\sin 3^0-\sin 3^0+.....=0+0+0+.....=0$

Hence, $\sin 1^0+\sin 2^0+\sin 3^0+.....+\sin {358}^0+\sin {359}^0=0$