Question

Among the given options $\cos {1^o} + \cos {2^o} + .... + \cos {180^o}$ is equal to
A.1
B.0
C.2
D.-1

Answer 1

In order to solve this question, You have to know the exact option for$\cos {1^o} + \cos {2^o} + .... + \cos {180^o}$. For that, first we should focus on the equation, whether the equation is additional or multiplication. Then, let's know the value of $\sin \theta$ and $\cos \theta$ in all $\theta$ .
The $\cos \theta$ is the ratio of the adjacent side to the hypotenuse, where $\theta$ is one of the acute angles.

Complete answer:
There are six trigonometric ratios sine, cosine, tangent, cosecant, secant and cotangent. These six trigonometric ratios are abbreviated as $\sin$ , $\cos$ , $\tan$ , $\cos ec$ , $\sec$ , $\cot$ . These are referred to as ratios since they can be expressed in terms of the sides of a right-angled triangle for a specific angle $\theta$.
Given equation $\cos {1^o} + \cos {2^o} + .... + \cos {180^o}$
Arrange the equation, as shown as below,
Let’s explains the equation up to cos180°
$\Rightarrow \cos {1^o} + \cos {2^o} + .... + \cos {180^o}$
$\Rightarrow cos{1^o} + cos{2^o} + cos{3^o} + .... + cos{90^o} + cos{91^o} + cos{92^o} + ...cos{180^o}$
The given equation can be rearranged as; its helps to simplify the equation simple and easy way,
$\Rightarrow \cos {1^o} + \cos {2^o} + ... + \cos {90^ \circ } + \cos {({180^o} - {89^o})^{}} + \cos ({180^o} - {88^o}) + ... + \cos ({180^o} - {0^o})$
But we know that: $\left( {\cos {{180}^o} - x} \right) = - \cos x$
Where $x$ is angle in degree.at the same time
$\Rightarrow \cos {90^ \circ } - x = \sin x$ Then,
$\Rightarrow (\cos {180^ \circ } + x) = - \cos x$
Let’s construct the following equation, it makes to solve easier
$\Rightarrow \left( {\cos {1^o} - \cos {1^o}} \right) + \left( {\cos {2^o} - \cos {2^o}} \right) + ....\left( {\cos {{89}^o} - \cos {{89}^o}} \right) + \left( {\cos {{90}^o} - \cos {0^o}} \right)$
So, all the bracket totals are to be zero, except last 2 terms will be zero. Because cos0=-1,thus the value of equation be
$\Rightarrow 0 + 0 + ...(0 - 1)$
We use $\cos 0 = - 1$
$\Rightarrow - 1$
Therefore the correct answer is option D .

Note:
This problem can be also solved by converting all the cosine angle terms into sine angle terms. After converting in sine terms we can proceed in the same way as it was done above but we need to take care of signs of sine because sine is positive in the first and second quadrant while it is negative in the third and fourth quadrant.