Recognize Limit Issue: First, let's recognize that as θ approaches 2π, sin(θ) approaches 1. This means that 1−sin2(θ) approaches 0, which could lead to a division by zero issue. We need to find a way to simplify the expression to avoid this.
Trigonometric Identity: We can use the trigonometric identity sin2(x)=1−cos2(x) to rewrite the denominator. This gives us:limθ→2π1−sin2(θ)sin2(2θ)=limθ→2πcos2(θ)sin2(2θ)
Double Angle Formula: Next, we can use the double angle formula for sine, which is sin(2x)=2sin(x)cos(x). Applying this to sin2(2θ), we get:sin2(2θ)=(2sin(θ)cos(θ))2=4sin2(θ)cos2(θ)
Substitute and Simplify: Substituting this back into our limit expression, we have:limθ→2πcos2(θ)4sin2(θ)cos2(θ)
Cancel Terms: We can now cancel out the cos2(θ) terms in the numerator and the denominator, which simplifies our expression to:limθ→2π4sin2(θ)
Evaluate Limit: As θ approaches 2π, sin(θ) approaches 1. Therefore, sin2(θ) approaches 12, which is 1. So our limit becomes:limθ→2π4sin2(θ)=4×1=4
Final Answer: The final answer is that the limit of sin2(2θ)/(1−sin2(θ)) as θ approaches π/2 is 4. Therefore, the correct choice is (C)4.
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