Question

Find exact values for sin θ and tan θ if cos θ = -√2/5 and θ terminates in Quadrant III.

Answer 1

Part 1)
`sin(\theta)=-(√(23))/(5)`

Part 2)
`tan(\theta)=(√(46))/(2)`

Explanation:

step 1

Find the
`sin(\theta)`

we know that

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

we have

`cos(\theta)=-(√(2))/(5)`

substitute

`sin^(2)(\theta) +(-(√(2))/(5))^(2)=1`

`sin^(2)(\theta) +(2)/(25)=1`

`sin^(2)(\theta)=1-(2)/(25)`

`sin^(2)(\theta)=(23)/(25)`

square root both sides

`sin(\theta)=\pm(√(23))/(5)`

Remember that the angle θ terminates in Quadrant III

That means, that the value of sin(θ) is negative

so

`sin(\theta)=-(√(23))/(5)`

step 2

Find the
`tan(\theta)`

we know that

`tan(\theta)=(sin(\theta))/(cos(\theta))`

we have

`sin(\theta)=-(√(23))/(5)`

`cos(\theta)=-(√(2))/(5)`

substitute

`tan(\theta)=-(√(23))/(5):-(√(2))/(5)=(√(23))/(√(2))`

simplify

`tan(\theta)=(√(46))/(2)`