mooltattawsmq8

2022-12-28

Find the value of cos ${225}^{\circ }$

eyrurin3wj

Expert

Determining the worth of $\mathrm{cos}{225}^{\circ }$
$\mathrm{cos}{225}^{\circ }$ can be expressed as
$\mathrm{cos}{225}^{\circ }=\mathrm{cos}\left({180}^{\circ }+{45}^{\circ }\right)$
$⇒$$\mathrm{cos}\left({180}^{\circ }+{45}^{\circ }\right)=\mathrm{cos}{180}^{\circ }×\mathrm{cos}{45}^{\circ }–\mathrm{sin}{180}^{\circ }×\mathrm{sin}{45}^{\circ }$ ; $\left[\mathrm{cos}\left(A+B\right)=\mathrm{cos}\left(A\right)×\mathrm{cos}\left(B\right)-\mathrm{sin}\left(A\right)×\mathrm{sin}\left(B\right)\right]$
When we substitute the above angles' values, we obtain,
$\mathrm{cos}{180}^{\circ }=-1,\mathrm{cos}{45}^{\circ }=\frac{2}{2},\mathrm{sin}{180}^{\circ }=0\mathrm{and}\mathrm{sin}{45}^{\circ }=\frac{2}{2}$
$\mathrm{cos}\left({180}^{\circ }+{45}^{\circ }\right)=-1×\left(\frac{2}{2}\right)-0×\left(\frac{2}{2}\right)=-\frac{2}{2}=-\frac{1}{2}$
Thus the value of $\mathrm{cos}{225}^{\circ }$ is $-\frac{1}{2}$

Do you have a similar question?