22. If
5
3
,cos
13
5
sin
u v
and u and v lie in QII, find the exact value of:
a)
sinu v
b)
cosu v
c)
secu v
d)
tanu v
To solve this problem, we need to use the sum and difference identities for trigonometric functions.
Given that sin(u) = 5/13 and cos(v) = 3/5, we can find cos(u) and sin(v) using the Pythagorean identity:
cos^2(u) + sin^2(u) = 1
(3/5)^2 + sin^2(u) = 1
9/25 + sin^2(u) = 1
sin^2(u) = 16/25
sin(u) = ±4/5
Since u lies in QII, sin(u) is positive. So sin(u) = 4/5.
Similarly, using the Pythagorean identity for cos(v), we can find sin(v):
cos^2(v) + sin^2(v) = 1
(5/13)^2 + sin^2(v) = 1
25/169 + sin^2(v) = 1
sin^2(v) = 144/169
sin(v) = ±12/13
Since v lies in QII, sin(v) is positive. So sin(v) = 12/13.
Now we can calculate the values of the given expressions:
a) sin(u + v) = sin(u)cos(v) + cos(u)sin(v)
= (4/5)(3/5) + (-3/5)(12/13)
= 12/25 - 36/65
= (312 - 900)/(25*13)
= -588/325
b) cos(u + v) = cos(u)cos(v) - sin(u)sin(v)
= (3/5)(3/5) - (4/5)(12/13)
= 9/25 - 48/65
= (585 - 960)/(25*13)
= -375/325
= -15/13
c) sec(u + v) = 1/cos(u + v)
= 1/(-15/13)
= -13/15
d) tan(u - v) = sin(u - v)/cos(u - v)
= (sin(u)cos(v) - cos(u)sin(v))/(cos(u)cos(v) + sin(u)sin(v))
= ((4/5)(3/5) - (-3/5)(12/13))/((3/5)(3/5) + (4/5)(12/13))
= (12/25 + 36/65)/(9/25 + 48/65)
= (312/325 + 936/169)/(225/325 + 480/169)
= (312*169 + 936*325)/(225*169 + 480*325)
= (52728 + 304200)/(38025 + 156000)
= 357/187