214k views
0 votes
The angle 0 lies in Quadrant III
cos 0 =1/6
What is tan
?

2 Answers

6 votes

Final answer:

To determine tan θ when cos θ = 1/6 and the angle lies in the third quadrant, you need to use the Pythagorean identity to find sin θ and then use the definition of tangent. The correct approach gives us tan θ = √35 / 6, considering that the values of sine and cosine are negative in the third quadrant.

Step-by-step explanation:

The question pertains to the function of tan θ in the context of trigonometry, specifically when the angle θ resides in the third quadrant and has a given cosine value. To determine the value of tan θ, we first have to understand that each trigonometric function has a specific sign associated with it depending on the quadrant the angle lies in. In the third quadrant, both sine and cosine are negative.

However, since the question provides us with cos θ = 1/6, we have to consider that the given value should be negative because cosine in the third quadrant is negative.

With cos θ known, we can find sin θ using the Pythagorean identity, sin^2 θ + cos^2 θ = 1. This gives us sin θ = -√(1 - cos^2 θ). Once we know both sine and cosine, we can use the definition of tangent, which is tan θ = sin θ / cos θ, to find our answer.

If cos θ = -1/6, then sin θ will be -√(1 - (-1/6)^2) = -√(35/36), and therefore, tan θ = sin θ / cos θ = -√(35/36) / (-1/6) = √35 / 6. This will be the value of tan θ when the angle is in the third quadrant with a cosine value of 1/6.

User KPheasey
by
4.5k points
5 votes

Answer:


tan(\theta)=√(35)

Step-by-step explanation:

we know that

The angle
\theta lies in Quadrant III

so

The value of cosine is negative

The value of sine is negative

The value of tangent is positive

we have


cos(\theta)=-(1)/(6)

step 1

Find the value of
sin(\theta)

Applying trigonometric identity


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

substitute the given value


sin^2(\theta)+(-(1)/(6))^2=1


sin^2(\theta)+(1)/(36)=1


sin^2(\theta)=1-(1)/(36)


sin^2(\theta)=(35)/(36)

square root both sides


sin(\theta)=\pm(√(35))/(6)

Remember that the sine is negative (Quadrant III)

so


sin(\theta)=-(√(35))/(6)

step 2

Find the value of
tan(\theta)

we know that


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

we have


sin(\theta)=-(√(35))/(6)


cos(\theta)=-(1)/(6)

substitute


tan(\theta)=-(√(35))/(6):-(1)/(6)=√(35)

User Krakkos
by
5.3k points