6. The Equation for the terminal Side `theta` is `y=1/6x` in Quadrant-3. Find value of all six trigonometric functions
Solution:
`y=1/6x`
`y=x/6`
In Quadrant-3, x is negative and y is negative
Let `x=-6`
So `y=1/6xx-6=-1`
`P(-6,-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`
Here `x=-6` and `y=-1`
In triangle ABC, by Pythagoras' theorem
`r^2 = x^2 + y^2`
`=(-6)^2 + (-1)^2`
`=36 + 1`
`=37`
`:.r=sqrt(37)`
So, `x=-6,y=-1 and r=sqrt(37)`
`(1)` `sin(theta)=y/r=(-1)/(sqrt(37))=(-sqrt(37))/37=-0.1644`
`(2)` `cos(theta)=x/r=(-6)/(sqrt(37))=(-6sqrt(37))/37=-0.98639`
`(3)` `tan(theta)=y/x=(-1)/(-6)=1/6=1/6`
`(4)` `csc(theta)=r/y=(sqrt(37))/(-1)=-sqrt(37)=-6.08276`
`(5)` `sec(theta)=r/x=(sqrt(37))/(-6)=(-sqrt(37))/6=-1.01379`
`(6)` `cot(theta)=x/y=(-6)/(-1)=6`
This material is intended as a summary. Use your textbook for detail explanation.
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