What is the tangent for an angle that has a cosine of −1/7√ and is in Quadrant III? Use the Pythagorean identity sin^2(θ)+cos^2(θ)=1 , the trigonometric identity tanθ=sinθ/cosθ , and the quadrant to solve. (1 point) Responses 6 6 −6–√ negative Start Root 6 End Root −6 negative 6 6–√
if cosθ = 1/√7, then sinθ = √6/√7
and tanθ = √6
since we are in QIII, tan is positive, so that is our answer.
Apologies for the incorrect calculation in my previous response.
You are correct. If cosθ = 1/√7, then sinθ = √6/√7, and tanθ = sinθ/cosθ = (√6/√7) / (1/√7) = √6.
Since we are in Quadrant III, where both sine and tangent are positive, the correct answer is √6.
Well done!
The given cosine is -1/7√, which can be simplified to -1√/7.
Since the cosine is negative in Quadrant III, we know that the sine will be positive.
Using the Pythagorean identity, sin²(θ) + cos²(θ) = 1, we can solve for the sine.
sin²(θ) + (-1√/7)² = 1
sin²(θ) + 1/7 = 1
sin²(θ) = 1 - 1/7
sin²(θ) = 6/7
Taking the square root of both sides, we get:
sin(θ) = √(6/7) = √6/√7 = √(6/7) * √(7/7) = √(42)/7
Now, we can use the tangent identity tanθ = sinθ/cosθ to find the tangent:
tanθ = (√(42)/7)/(-1√/7) = -√42/-1 = √42
Therefore, the tangent for an angle that has a cosine of -1/7√ and is in Quadrant III is √42.
The correct answer is:
√42