Sunday, July 24, 2016

`csc(u - v)` 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`


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


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


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


since u is in Quadrant III ,


`cos(u)=-24/25`


Now `sin^2(v)+cos^2(v)=1`  


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


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


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


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


since v is in Quadrant III , 


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


`csc(u-v)=1/(sin(u-v))`


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


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


`=1/(28/125-72/125)`


`=-125/44`

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