Question 1032545
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If sin α = 4/5 and cos β = -5/13 for α in Quadrant I and β in Quadrant II, find sin(α - β) and cos(α - β)
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Use the formulas 

{{{sin(alpha-beta)}}} = {{{sin(alpha)*cos(beta) - cos(alpha)*sin(beta)}}}    (1)

   and

{{{cos(alpha-beta)}}} = {{{cos(alpha)*cos(beta) + sin(alpha)*sin(beta)}}}.   (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(alpha)}}} = {{{4/5}}}  and  {{{cos(beta)}}} = {{{-5/13}}}, you need to know  {{{cos(alpha)}}} and {{{sin(beta)}}}.


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

   The sign "+" was chosen for the square root because {{{alpha}}} is in Q1.


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

   The sign "+" was chosen for the square root because {{{beta}}} is in Q2.


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


{{{sin(alpha-beta)}}} = {{{(4/5)*(-5/13) - (3/5)*(12/13)}}} = {{{-20/65 - 36/65}}} = {{{(-20-36)/65}}} = {{{-56/65}}},   and

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