Q.94 - Precalculus Enhanced with Graphing Utilities

Question

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

Step-by-Step Solution

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 !: Given information

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

Step 2: Use the periodicity of the cosine function cos 180 - θ = - cos θ cos 360 - θ = cos θ

We get

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

Step 3: Conclusion

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