Use the given sum or difference identity to find cos⁡(α−β)\displaystyle {\cos{{\left(\alpha-\beta\right)}}}  exactly, given that sin⁡α=−47\displaystyle {\sin{\alpha}}=-\frac{{4}}{{7}} such that α\displaystyle \alpha terminates in quadrant III and cos⁡β=17\displaystyle {\cos{\beta}}=\frac{{1}}{{7}} such that β\displaystyle \beta terminates in quadrant IV then:

 cos⁡(α−β)=cos⁡(α)cos⁡(β)+sin⁡(α)sin⁡(β)\displaystyle {\cos{{\left(\alpha-\beta\right)}}}={\cos{{\left(\alpha\right)}}}{\cos{{\left(\beta\right)}}}+{\sin{{\left(\alpha\right)}}}{\sin{{\left(\beta\right)}}} 

 cos⁡(α−β)=(\displaystyle {\cos{{\left(\alpha-\beta\right)}}}={(} )\displaystyle {)}(17)+\displaystyle {\left(\frac{{1}}{{7}}\right)}+(−47)(\displaystyle {\left(-\frac{{4}}{{7}}\right)}{(} )\displaystyle {)} 

 cos⁡(α−β)=\displaystyle {\cos{{\left(\alpha-\beta\right)}}}=

Your answers should be exact but you do not have to rationalize your denominators.

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