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Let V = Ÿ¨sinθ, cosθ, 8sinθ + 3cos(2θ)Ÿ©. Find the projection of V onto the xy-plane for -1 ≤ θ ≤ 1.

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Final answer:

To find the projection of V onto the xy-plane, we set the z-coordinate of V to zero by solving the equation 8sinθ + 3cos(2θ) = 0 using trigonometric identities. The solutions will give us the values of θ for the projection.

Step-by-step explanation:

To find the projection of V onto the xy-plane, we need to set the z-coordinate of V to zero. In other words, we need to find the values of θ that satisfy the equation 8sinθ + 3cos(2θ) = 0.

To solve this equation, we can use the trigonometric identities sin2θ = 1 - cos2θ and cos(2θ) = 2cos2θ - 1. Substituting these identities into the equation, we get 8sinθ + 3(2cos2θ - 1) = 0.

From here, we can simplify the equation and solve for θ. The solutions will give us the values of θ for which the z-coordinate of V is zero, and thus, the projection of V onto the xy-plane.

User Mike Hogan
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