Question
What is the sine for an angle that has a cosine of −47 and is in Quadrant II? Use the Pythagorean identity sin2(θ)+cos2(θ)=1 and the quadrant to solve.(1 point) Responses −33√7 negative Start Fraction Start Root 33 End Root over 7 End Fraction 33√7 Start Fraction Start Root 33 End Root over 7 End Fraction −3349 negative Start Fraction 33 over 49 End Fraction 3349
Answers
GPT 3.5
We know that the cosine of an angle in Quadrant II is negative. Since the cosine is -47, we can use the Pythagorean identity sin^2(θ) + cos^2(θ) = 1 to find the value of sin(θ).
Plugging in the value of cos(θ) = -47 into the identity, we get:
sin^2(θ) + (-47)^2 = 1
sin^2(θ) + 2209 = 1
sin^2(θ) = -2208
Since the sine is positive in Quadrant II, we can take the positive square root:
sin(θ) = √(-2208)
sin(θ) = √(-1) * √(2208)
sin(θ) = √(-1) * √(16 * 138)
sin(θ) = 4√(-1) * √(138)
sin(θ) = 4i√(138)
Therefore, the sine for the angle with cosine -47 in Quadrant II is 4i√(138), where i is the imaginary unit.
Plugging in the value of cos(θ) = -47 into the identity, we get:
sin^2(θ) + (-47)^2 = 1
sin^2(θ) + 2209 = 1
sin^2(θ) = -2208
Since the sine is positive in Quadrant II, we can take the positive square root:
sin(θ) = √(-2208)
sin(θ) = √(-1) * √(2208)
sin(θ) = √(-1) * √(16 * 138)
sin(θ) = 4√(-1) * √(138)
sin(θ) = 4i√(138)
Therefore, the sine for the angle with cosine -47 in Quadrant II is 4i√(138), where i is the imaginary unit.