2. If `cos(x)=12/13`, find other trigonometry functions `sin(x),cos(x),tan(x),csc(x),sec(x),cot(x)`
Solution:
`cos(x)=12/13`, in Quadrant-1
Opposite side `(y)`, adjacent side `(x)` and hypotenuse `(r)`
`sin(theta), cos(theta), tan(theta)` fromula
`sin(theta) = "opposite"/"hypotenuse" = y/r`
`cos(theta) = "adjacent"/"hypotenuse" = x/r`
`tan(theta) = "opposite"/"adjacent" = y/x`
`csc(theta) = "hypotenuse"/"opposite" = r/y`
`sec(theta) = "hypotenuse"/"adjacent" = r/x`
`cot(theta) = "adjacent"/"opposite" = x/y`
`cos(x) = "adjacent"/"hypotenuse" = x/r = 12/13`
Here `x=12` and `r=13`
In triangle ABC, by Pythagoras' theorem
`r^2=x^2+y^2`
`:.y^2=r^2-x^2`
`=13^2-12^2`
`=169-144`
`=25`
`:.y=sqrt(25)=5` (`:'` y is +ve in Quadrant-1)
So, `x=12,y=5 and r=13`
`(1)` `sin(x)=y/r=(5)/(13)=5/13`
`(2)` `cos(x)=x/r=(12)/(13)=12/13`
`(3)` `tan(x)=y/x=(5)/(12)=5/12`
`(4)` `csc(x)=r/y=(13)/(5)=13/5`
`(5)` `sec(x)=r/x=(13)/(12)=13/12`
`(6)` `cot(x)=x/y=(12)/(5)=12/5`
Second Method
`cos(x)=12/13`, in Quadrant-1
`(1)` `sin^2(x)=1-cos^2(x)`
`=1-(12/13)^2`
`=1-144/169`
`=(169-144)/169`
`=25/169`
`:.sin(x)=sqrt(25/169)=5/13=5/13`
`(2)` `tan(x)=sin(x)/cos(x)=(5/13)/(12/13)=5/13 xx 13/12=5/12=5/12`
`(3)` `sec(x)=1/cos(x)=1/(12/13)=13/12=13/12`
`(4)` `csc(x)=1/sin(x)=1/(5/13)=13/5=13/5`
`(5)` `cot(x)=1/tan(x)=1/(5/12)=12/5=12/5`
This material is intended as a summary. Use your textbook for detail explanation.
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