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Question

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

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Solution

## Evaluating the value of $\mathrm{cos}{225}^{\circ }$$\mathrm{cos}{225}^{\circ }$ can be expressed as $\mathrm{cos}{225}^{\circ }=\mathrm{cos}\left({180}^{\circ }+{45}^{\circ }\right)$$⇒$$\mathrm{cos}\left({180}^{\circ }+{45}^{\circ }\right)=\mathrm{cos}{180}^{\circ }×\mathrm{cos}{45}^{\circ }–\mathrm{sin}{180}^{\circ }×\mathrm{sin}{45}^{\circ }$ ; $\left[\mathrm{cos}\left(A+B\right)=\mathrm{cos}\left(A\right)×\mathrm{cos}\left(B\right)-\mathrm{sin}\left(A\right)×\mathrm{sin}\left(B\right)\right]$Substituting the values of the above angles we get,$\mathrm{cos}{180}^{\circ }=-1,\mathrm{cos}{45}^{\circ }=\frac{\sqrt{2}}{2},\mathrm{sin}{180}^{\circ }=0\mathrm{and}\mathrm{sin}{45}^{\circ }=\frac{\sqrt{2}}{2}$$\begin{array}{rcl}\mathrm{cos}\left({180}^{\circ }+{45}^{\circ }\right)& =& -1×\left(\frac{\sqrt{2}}{2}\right)-0×\left(\frac{\sqrt{2}}{2}\right)\\ & =& -\frac{\sqrt{2}}{2}\\ & =& -\frac{1}{\sqrt{2}}\\ & & \end{array}$Hence the value of $\mathrm{cos}{225}^{\circ }$ is $-\frac{1}{\sqrt{2}}$

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