Question

If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ?

Answer 1

For the answer to the question above, If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ?

If 3 pi over 2 < Θ < 2π Then the signs of sinx and tanx are both negative.

cos(x) = sqrt(2)/2. So sin(x) = -sqrt(2)/2 and tan(x) = -1.

sin Θ = square root 2 over 2; tan Θ = −1

sin Θ = negative square root 2 over 2; tan Θ = 1

sin Θ = square root 2 over 2; tan Θ = negative square root 2

sin Θ = negative square root 2 over 2; tan Θ = −1

Answer 2

The answer is

`sin(\theta)=-(√(2))/(2)`

`tan(\theta)=-1`

Explanation:

we know that

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

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

In this problem we have

`cos(\theta)=(√(2))/(2)`

`(3\pi)/(2) <\theta < 2\pi`

so

The angle
`\theta` belong to the third or fourth quadrant

The value of
`sin(\theta)` is negative

Step 1

Find the value of
`sin(\theta)`

Remember

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

we have

`cos(\theta)=(√(2))/(2)`

substitute

`sin^(2)(\theta)+((√(2))/(2))^(2)=1`

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

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

`sin(\theta)=-(√(2))/(2)` ------> remember that the value is negative

Step 2

Find the value of
`tan(\theta)`

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

we have

`sin(\theta)=-(√(2))/(2)`

`cos(\theta)=(√(2))/(2)`

substitute

`tan(\theta)=(-(√(2))/(2))/((√(2))/(2))`

`tan(\theta)=-1`