Use the given sum or difference identity to find cos⁡(α−β)\displaystyle {\cos{{\left(\alpha-\beta\right)}}}  exactly, given that cos⁡α=−34\displaystyle {\cos{\alpha}}=-\frac{{3}}{{4}} such that α\displaystyle \alpha terminates in quadrant III and cos⁡β=−13\displaystyle {\cos{\beta}}=-\frac{{1}}{{3}} such that β\displaystyle \beta terminates in quadrant II 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)=(−34)(−13)+(\displaystyle {\cos{{\left(\alpha-\beta\right)}}}={\left(-\frac{{3}}{{4}}\right)}{\left(-\frac{{1}}{{3}}\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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