Question 1039196
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use the cosine of a sum and cosine of a difference identities to find cos(s+t) and cos(s-t).

sin s=  -4/5  in Q IV
sin t=  12/13 in Q II

what is cos(s+t)?
what is cos(s-t)?

thank you in advance!!!
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Use the formulas 

cos(s+t) = cos(s)*cos(t) - sin(s)*sin(t)    (1)

   and

cos(s-t) = cos(s)*cos(t) + sin(s)*sin(t)    (2)


Regarding these formulas, see the lesson <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Addition-and-subtraction-formulas.lesson>Addition and subtraction formulas</A> in this site.


In addition to the given  sin(s) = {{{-4/5}}}  and  sin(t) = {{{12/13}}}, you need to know  cos(s) and cos(t).


1.  cos(s) = {{{sqrt(1-sin^2(s))}}} = {{{sqrt(1 - (-4/5)^2)}}} = {{{sqrt(1 - 16/25)}}} = {{{sqrt(25-16)/25)}}} = {{{sqrt(9/25)}}} = {{{3/5}}}.

   The sign "+" was chosen for the square root because cos(s) is positive in Q4.


2.  cos(t) = = -{{{sqrt(1-sin^2(t))}}} = -{{{sqrt(1 - (12/13)^2)}}} = -{{{sqrt(1 - 144/169)}}} = -{{{sqrt(169-144)/169)}}} = -{{{sqrt(25/169)}}} ={{{-5/13}}}.

   The sign "-" was chosen for the square root because cos(t) is negative in Q2.


Now all you need to do is to substitute everything into the formulas (1) and (2) and make the calculations.


cos(s-t) = {{{(3/5)*(-5/13) + (-4/5)*(12/13)}}} = {{{-15/65 - 48/65}}} = {{{(-15-48)/65}}} = {{{-63/65}}},   and

cos(s+t) = {{{(3/5)*(-5/13) - (-4/5)*(12/13)}}} = {{{-15/65 + 48/65}}} = {{{(-15+48)/65}}} = {{{33/65}}}.
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Solved.