Question 1099718
{{{cos(x+y)=cos(x)cos(y)-sin(x)sin(y)}}}
and
{{{cos^2(x)+sin^2(x)=1}}}
So then,
{{{cos^2(alpha)+sin^2(alpha)=1}}}
{{{cos^2(alpha)+9/25=1}}}
{{{cos^2(alpha)=16/25}}}
Since {{{alpha}}} is in Q3, cosine is negative so,
{{{cos(alpha)=-4/5}}}
and similarly,
{{{cos^2(beta)+25/169=1}}}
{{{cos^2(beta)=144/169}}}
Since {{{beta}}} is in Q1, cosine is positive so,
{{{cos(beta)=12/13}}}
So putting it all together,
{{{cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta)}}}
{{{cos(alpha+beta)=(-4/5)(12/13)-(-3/5)(5/13)}}}
Work through that to get the answer.