Wednesday, December 31, 2014

`sec(v - u)` Find the exact value of the trigonometric expression given that sin(u) = -7/25 and cos(v) = -4/5 (Both u and v are in quadrant III.)

Given `sin(u)=-7/25, cos(v)=-4/5`


using pythagorean identity,


`sin^2(u)+cos^2(u)=1`


`(-7/25)^2+cos^2(u)=1`


`cos^2(u)=1-49/625=(625-49)/625=576/625`


`cos(u)=sqrt(576/625)`


`cos(u)=+-24/25`


since u is in quadrant III,


`:.cos(u)=-24/25`


`sin^2(v)+cos^2(v)=1`


`sin^2(v)+(-4/5)^2=1`


`sin^2(v)+16/25=1`


`sin^2(v)=1-16/25=(25-16)/25=9/25`


`sin(v)=sqrt(9/25)`


`sin(v)=+-3/5`


since v is in quadrant III,


`:.sin(v)=-3/5`


Now let's evaluate sec(v-u),


`sec(v-u)=1/cos(v-u)`


`=1/(cos(v)cos(u)+sin(v)sin(u))`


`=1/((-4/5)(-24/25)+(-3/5)(-7/25))`


`=1/(96/125+21/125)`


`=125/117`

No comments:

Post a Comment