1. If `sin(x)=3/5`, find other trigonometry functions `sin(x),cos(x),tan(x),csc(x),sec(x),cot(x)`
Solution:
`sin(x)=3/5`, 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`
`sin(x) = "opposite"/"hypotenuse" = y/r = 3/5`
Here `y=3` and `r=5`
In triangle ABC, by Pythagoras' theorem
`r^2=x^2+y^2`
`:.x^2=r^2-y^2`
`=5^2-3^2`
`=25-9`
`=16`
`:.x=sqrt(16)=4` (`:'` x is +ve in Quadrant-1)
So, `x=4,y=3 and r=5`
`(1)` `sin(x)=y/r=(3)/(5)=3/5`
`(2)` `cos(x)=x/r=(4)/(5)=4/5`
`(3)` `tan(x)=y/x=(3)/(4)=3/4`
`(4)` `csc(x)=r/y=(5)/(3)=5/3`
`(5)` `sec(x)=r/x=(5)/(4)=5/4`
`(6)` `cot(x)=x/y=(4)/(3)=4/3`
Second Method
`sin(x)=3/5`, in Quadrant-1
`(1)` `cos^2(x)=1-sin^2(x)`
`=1-(3/5)^2`
`=1-9/25`
`=(25-9)/25`
`=16/25`
`:.cos(x)=sqrt(16/25)=4/5=4/5`
`(2)` `tan(x)=sin(x)/cos(x)=(3/5)/(4/5)=3/5 xx 5/4=3/4=3/4`
`(3)` `csc(x)=1/sin(x)=1/(3/5)=5/3=5/3`
`(4)` `sec(x)=1/cos(x)=1/(4/5)=5/4=5/4`
`(5)` `cot(x)=1/tan(x)=1/(3/4)=4/3=4/3`
This material is intended as a summary. Use your textbook for detail explanation.
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