x and a are each acute angles in standard position. sinx = 3/5 and cosb = 12/13. determine cos(b+x)
To find cos(b+x), we can use the trigonometric identity:
cos(b+x) = cos(b)cos(x) - sin(b)sin(x)
We are given that cos(b) = 12/13 and sin(x) = 3/5.
To find sin(b), we can use the Pythagorean identity:
sin^2(b) + cos^2(b) = 1
sin^2(b) = 1 - cos^2(b)
sin(b) = √(1 - cos^2(b))
sin(b) = √(1 - (12/13)^2)
sin(b) = √(1 - 144/169)
sin(b) = √(169/169 - 144/169)
sin(b) = √(25/169)
sin(b) = 5/13
Now, we can substitute the values into the formula:
cos(b+x) = cos(b)cos(x) - sin(b)sin(x)
cos(b+x) = (12/13) * cos(x) - (5/13) * (3/5)
cos(b+x) = (12/13) * cos(x) - (15/13)
cos(b+x) = (12/13) * (4/5) - (15/13)
cos(b+x) = 48/65 - 15/13
cos(b+x) = (48/65) - (15/13)
cos(b+x) = (48*13 - 65*15)/(65*13)
cos(b+x) = (624 - 975)/845
cos(b+x) = -351/845
Therefore, cos(b+x) = -351/845.