Question

Let (-5, 6) be a point on the terminal side of o.

Find the exact values of sin 0, cscO, and coto.

Answer 1

The exact values of sin
`\theta` , csc
`\theta` , and cot
`\theta` are:

sin
`\theta` =
`\frac {6} \sqrt(61)`

csc
`\theta` =
`\frac {\sqrt(61)} { 6}`

cot
`\theta` = -5 / 6

How to find the exact values

To find the exact values of sin
`\theta` , csc
`\theta` , and cot
`\theta` , determine the values of the trigonometric ratios based on the given point (-5, 6) on the terminal side of
`\theta` .

Using the given coordinates, calculate the radius (r) and the angle (
`\theta` ) using the formulas:

`r = \sqrt((-5)^2 + 6^2) \\= \sqrt(25 + 36) \\= \sqrt(61)`

`\theta` = arctan(y/x) = arctan(6/(-5))

Let's calculate the values:

sin
`\theta` = y / r =
`\frac {6} \sqrt(61)`

csc
`\theta` = 1 / sin
`\theta` =
`\frac {\sqrt(61)} { 6}`

cot
`\theta` = x / y = -5 / 6

Therefore, the exact values of sin
`\theta` , csc
`\theta` , and cot
`\theta` are:

sin
`\theta` =
`\frac {6} \sqrt(61)`

csc
`\theta` =
`\frac {\sqrt(61)} { 6}`

cot
`\theta` = -5 / 6