Question

Q5 Q6.) Find the exact value of the six trigonometric functions of theta.

Answer 1

two pts given (-2, -5) n (0,0)
hypothesis is line segment bet 2 pts
= sqrt (-2^2 + -5^2)
= sqrt (4 + 25)
= sqrt (29)

sin
`\theta` = y/hypo = -5 / sqrt (29)

cos
`\theta` = x/hypo = -2 / sqrt (29)

tan
`\theta` = y/x = -5 / -2 = 2 1/2

csc
`\theta` = hypo/y = -sqrt (29) / 5

sec
`\theta` = hypo/x = -sqrt (29) / 2

cot
`\theta` = x/y = -2 / -5 = 2 / 5

Answer 2

5y - 2x + 1 = 0.
5y = 2x-1.

y = 2/5 x - 2/5

this tells us that tan(theta) = 2/5
sin(theta)/sqrt(1-sin^2(theta)) = 2/5
x/sqrt(1-x^2)=2/5
x^2/(1-x^2)=4/25
25x^2=4-4x^2
29x^2=4
x^2=4/29
x = -2/sqrt(29) because quadrant III
cos(x) = sqrt(1-4/29) = -5/sqrt(29)


tan(theta)=2/5
cos(theta)=-5/sqrt(29)
sin(theta)=-2/sqrt(29)
cot(theta)=5/2
sec(theta)=-sqrt(29)/5
csc(theta)=-sqrt(29)/2