Question 1171626
cos(a) =-4/5 and a is in quadrant II and sin(b)=5/13 and b is in quadrant II. Find cos(a+b)
<pre>cos (a + b) = cos (a) cos (b) - sin (a) sin (b)
From above, it's obvious that we need cos (b) and sin (a).
Given {{{matrix(1,3, cos (a), "=", (- 4)/5)}}}, and that &#8737a is in the 2nd quadrant, we should realize that we're dealing with a 3-4-5 PYTHAGOREAN TRIPLE, which means that {{{matrix(1,3, sin (a), "=", 3/5)}}}.
Likewise, given {{{matrix(1,3, sin (b), "=", 5/13)}}}, and that &#8737b is in the 2nd quadrant, we should realize that we're dealing with a 5-12-13 PYTHAGOREAN TRIPLE, which means that {{{matrix(1,3, cos (b), "=", (- 12)/13)}}}.
We now have: {{{highlight_green(system(matrix(3,3, cos (a + b), "=", cos (a) cos (b) - sin (a) sin (b),
cos (a + b), "=", ((- 4)/5) * ((- 12)/13) - (3/5) * (5/13),
cos (a + b), "=", 48/65 - 15/65),
highlight(matrix(1,3, cos (a + b), "=", 33/65))))}}}