Question 868504
From trig relation,
{{{cos(a+b)=cos(a)cos(b)-sin(a)sin(b)}}}
So you need to find {{{cos(a)}}} and {{{cos(b)}}}.
Since {{{cos^2(a)+sin^2(a)=1}}}
then,
{{{cos^2(a)+(4/5)^2=1}}}
{{{cos^2(a)=25/25-16/25}}}
{{{cos^2(a)=9/25}}}
{{{cos(a)=0 +- 3/5}}}
Since you're given information that {{{pi/2<a<pi}}}, then {{{cos(a)}}} must be negative (2nd quadrant) so you choose,
{{{cos(a)=-3/5}}}
Now for {{{b}}}.
Same thing,
{{{cos^2(b)+sin^2(b)=1}}}
{{{cos^2(b)+(-(2sqrt(5))/5)^2=1}}}
{{{cos^2(b)+(4/5)=1}}}
{{{cos^2(b)=5/5-4/5}}}
{{{cos^2(b)=1/5}}}
{{{cos(b)=0 +- sqrt(5)/5}}}
From the information, {{{b}}} lies in the 3rd quadrant, where the cosine is also negative
{{{cos(b)=-sqrt(5)/5}}}
Now go back to the original equation and plug in the values.
{{{cos(a+b)=cos(a)cos(b)-sin(a)sin(b)}}}
{{{cos(a+b)=(-3/5)(-sqrt(5)/5))-(4/5)(-(2sqrt(5))/5)}}}
{{{cos(a+b)=(3sqrt(5))/25+(8sqrt(5))/25)}}}
{{{cos(a+b)=(11sqrt(5))/25}}}
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As a check using inverse trig functions,
{{{a=126.9}}} degrees
{{{b=243.4}}} degrees
{{{a+b=370.3}}} degrees
{{{cos(370.3)=0.9839}}}
{{{11*sqrt(5)/25=0.9839}}}