Find the exact value of the trigonometric expression given that sin(u) = 21/ 29 and cos(v) = − 15 /17 . (Both u and v are in Quadrant II.)

Question

cos(v − u)

= _____________

Answer 1

Let "h" be the hypotenuse of the right triangle with angle "u or angle "v" in the unit-circle.

Let "x", "y" be the legs of the right triangle with angle "u" or angle "v" in the unit-circle.

sin(u) = 21/29

y = 21 (units), h = 29 (units)

x = √29² - 21² = 20 (units)

Calculating cos(u):

cos(u) = x/h = -20/29 ← angle "u" is located in Quadrant II, so "x" is negative and cos(u) is negative

cos(v) = -15/17

x = 15 (units), h = 17 (units)

y = √17² - 15² = 8 (units)

sin(v) = y/h = 8/17

cos(v) = -15/17 sin(v) = 8/17

cos(u) = -20/29 sin(u) = 21/29

cos(v - u) = cos(v) • cos(u) + sin(v) • sin(u)

cos (v - u) = (-15/17)(-20/29) + (8/17)(21/29) = 300/493 + 168/493 = 468/493

Answer 2

identity equation for cosine of the difference of two angles: cos(v-u) = cos(v)cos(u) + sin(v)sin(u)

use identity sin(u)^2 + cos(u)^2 = 1 and given value of sin(u) to find cos(u), cos(u) will be negative as in QII.

use identity sin(v)^2 + cos(v)^2 = 1 and given value of cos(v) to find sin(v), sin(v) will be positive as in QII.

plug given and computed values into identity equation.