Question

Find the exact value of$\cos \left(1^0\right)+\cos \left(2^o\right)+\cos \left(3^o\right)+.....+\cos \left({359}^o\right)$

Short answer

The exact value of$\cos \left(1^0\right)+\cos \left(2^o\right)+\cos \left(3^o\right)+.....+\cos \left({359}^o\right)is-1$

Step-by-step solution

Step !: Given information

We are given a equation$\cos \left(1^0\right)+\cos \left(2^o\right)+\cos \left(3^o\right)+.....+\cos \left({359}^o\right)$

Use the periodicity of the cosine function cos180-θ=-cosθcos360-θ=cosθ

We get

$$\begin{gathered} \cos \left(1^0\right)+\cos \left(2^o\right)+\cos \left(3^o\right)+...+\cos \left({358}^o\right)+\cos \left({359}^o\right)+\cos \left({360}^o\right) \\ =0-1 \\ =-1 \end{gathered}$$

Conclusion

The value of$\cos \left(1^0\right)+\cos \left(2^o\right)+\cos \left(3^o\right)+...+\cos \left({360}^o\right)is-1$