Respuesta :
Remember that
[tex]cos(\alpha-\beta)=cos\alpha *cos\beta+sin\alpha *sin\beta[/tex]step 1
Find out the cosine of angle alpha
Remember that
[tex]\begin{gathered} sin^2\alpha+cos^2\alpha=1 \\ sin\alpha\text{=}\frac{3}{5}\text{ ---> given} \end{gathered}[/tex]substitute
[tex](\frac{3}{5})^2+cos^2\alpha=1[/tex]The angle alpha lies on the II quadrant -----> the value of the cosine is negative
[tex]\begin{gathered} cos^2\alpha=1-\frac{9}{25} \\ cos^2\alpha=\frac{16}{25} \\ cos^\alpha=-\frac{4}{5} \end{gathered}[/tex]step 2
Find out the sine of the angle beta
[tex]\begin{gathered} s\imaginaryI n^2\beta+cos^2\beta=1 \\ cos\beta=-\frac{12}{13} \end{gathered}[/tex]substitute
[tex]s\imaginaryI n^2\beta+(-\frac{12}{13})^2=1[/tex]the angle beta lies on the III Quadrant -----> the value of the sine is negative
[tex]\begin{gathered} s\imaginaryI n^2\beta=1-\frac{144}{169} \\ s\mathrm{i}n^2\beta=\frac{25}{169} \\ s\mathrm{i}n\beta=-\frac{5}{13} \end{gathered}[/tex]step 3
Substitute the given values in the formula of step 1
[tex]\begin{gathered} cos(\alpha-\beta)=cos\alpha cos\beta+s\imaginaryI n\alpha s\imaginaryI n\beta \\ cos(\alpha-\beta)=(-\frac{4}{5})(-\frac{12}{13})+(\frac{3}{5})(-\frac{5}{13}) \end{gathered}[/tex][tex]\begin{gathered} cos(\alpha-\beta)=\frac{48}{65}-\frac{15}{65} \\ \\ cos(\alpha-\beta)=\frac{33}{65} \end{gathered}[/tex]