Answer:
[tex]cos(\alpha+\beta)=\frac{33}{65}[/tex]
Step-by-step explanation:
step 1
Find cos α
we know that
[tex]tan^2(\alpha)+1=sec^2(\alpha)[/tex]
we have
[tex]tan(\alpha)=-\frac{12}{5}[/tex]
substitute
[tex](-\frac{12}{5})^2+1=sec^2(\alpha)[/tex]
[tex]sec^2(\alpha)=\frac{144}{25}+1[/tex]
[tex]sec^2(\alpha)=\frac{169}{25}[/tex]
[tex]sec(\alpha)=\pm\frac{13}{5}[/tex]
Remember that Angle α lies in quadrant II
so
sec α is negative
[tex]sec(\alpha)=-\frac{13}{5}[/tex]
Find the value of cos α
[tex]cos)\alpha)=\frac{1}{sec(\alpha)}[/tex]
so
[tex]cos(\alpha)=-\frac{5}{13}[/tex]
step 2
Find sin α
we know that
[tex]tan(\alpha)=\frac{sin(\alpha)}{cos(\alpha)}[/tex]
[tex]sin(\alpha)=tan(\alpha)cos(\alpha)[/tex]
we have
[tex]tan(\alpha)=-\frac{12}{5}[/tex]
[tex]cos(\alpha)=-\frac{5}{13}[/tex]
substitute
[tex]sin(\alpha)=(-\frac{12}{5})(-\frac{5}{13})[/tex]
[tex]sin(\alpha)=\frac{12}{13}[/tex]
step 3
Find sin β
we know that
[tex]sin^2(\beta)+cos^2(\beta)=1[/tex]
we have
[tex]cos(\beta)=\frac{3}{5}[/tex]
substitute
[tex]sin^2(\beta)+(\frac{3}{5})^2=1[/tex]
[tex]sin^2(\beta)=1-(\frac{3}{5})^2[/tex]
[tex]sin^2(\beta)=1-\frac{9}{25}[/tex]
[tex]sin^2(\beta)=\frac{16}{25}[/tex]
[tex]sin(\beta)=\pm\frac{4}{5}[/tex]
Remember that
Angle β lies in quadrant IV
so
sin β is negative
[tex]sin(\beta)=-\frac{4}{5}[/tex]
step 4
Find cos(α−β)
we know that
[tex]cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)[/tex]
we have
[tex]cos(\alpha)=-\frac{5}{13}[/tex]
[tex]cos(\beta)=\frac{3}{5}[/tex]
[tex]sin(\alpha)=\frac{12}{13}[/tex]
[tex]sin(\beta)=-\frac{4}{5}[/tex]
substitute the given values
[tex]cos(\alpha+\beta)=(-\frac{5}{13})(\frac{3}{5})-(\frac{12}{13})(-\frac{4}{5})[/tex]
[tex]cos(\alpha+\beta)=(-\frac{15}{65})+(\frac{48}{65})[/tex]
[tex]cos(\alpha+\beta)=\frac{33}{65}[/tex]
