Question

Suppose θ is an angle in the standard position whose terminal side is in Quadrant IV and cot θ= -6/7 . Find the exact values of the five remaining trigonometric functions of θ. Show your work

Answer 1

Part 1)
`csc(\theta)=-(√(85))/(7)`

Part 2)
`sin(\theta)=-(7)/(√(85))` or
`sin(\theta)=-(7√(85))/(85)`

Part 3)
`tan(\theta)=-(7)/(6)`

Part 4)
`cos(\theta)=(6)/(√(85))` or
`cos(\theta)=(6√(85))/(85)`

Part 5)
`sec(\theta)=(√(85))/(6)`

Explanation:

we know that

The angle theta lie on the IV Quadrant

so

sin(θ) is negative

cos(θ) is positive

tan(θ) is negative

sec(θ) is positive

csc(θ) is negative

step 1

Find the value of csc(θ)

we know that

`1+cot^(2)(\theta)=csc^(2)(\theta)`

we have

`cot(\theta)=-(6)/(7)`

substitute

`1+(-(6)/(7))^(2)=csc^(2)(\theta)`

`1+(36)/(49)=csc^(2)(\theta)`

`(85)/(49)=csc^(2)(\theta)`rewrite

`csc(\theta)=-(√(85))/(7)` ----> remember that is negative

step 2

Find the value of sin(θ)

we know that

`csc(\theta)=(1)/(sin(\theta))`

we have

`csc(\theta)=-(√(85))/(7)`

therefore

`sin(\theta)=-(7)/(√(85))`

or

`sin(\theta)=-(7√(85))/(85)`

step 3

Find the value of tan(θ)

we know that

`tan(\theta)=(1)/(cot(\theta))`

we have

`cot(\theta)=-(6)/(7)`

therefore

`tan(\theta)=-(7)/(6)`

step 4

Find the value of cos(θ)

we know that

`sin^(2)(\theta)+cos^(2)(\theta)=1`

we have

`sin(\theta)=-(7)/(√(85))`

substitute

`(-(7)/(√(85)))^(2)+cos^(2)(\theta)=1`

`(49)/(85)+cos^(2)(\theta)=1`

`cos^(2)(\theta)=1-(49)/(85)`

`cos^(2)(\theta)=(36)/(85)`

`cos(\theta)=(6)/(√(85))` ------> the cosine is positive

or

`cos(\theta)=(6√(85))/(85)`

step 5

Find the value of sec(θ)

we know that

`sec(\theta)=(1)/(cos(\theta))`

we have

`cos(\theta)=(6)/(√(85))`

therefore

`sec(\theta)=(√(85))/(6)` ----> is positive