review the proof. which step contains an error? step statement 1 cos(2x) = 1 - 2sin²(x) 2 let 2x = θ. 3 then…

review the proof. which step contains an error? step statement 1 cos(2x) = 1 - 2sin²(x) 2 let 2x = θ. 3 then x = θ/2. 4 cos(θ) = 1 - 2sin²(θ/2) 5 -1 + cos(θ) = -2sin²(θ/2) 6 1 + cos(θ) = 2sin²(θ/2) 7 (1 - cos(θ))/2 = sin²(θ/2) 8 sin(θ/2) = ±√((1 - cos(θ))/2) step 2 step 4 step 6 step 8

review the proof. which step contains an error? step statement 1 cos(2x) = 1 - 2sin²(x) 2 let 2x = θ. 3 then x = θ/2. 4 cos(θ) = 1 - 2sin²(θ/2) 5 -1 + cos(θ) = -2sin²(θ/2) 6 1 + cos(θ) = 2sin²(θ/2) 7 (1 - cos(θ))/2 = sin²(θ/2) 8 sin(θ/2) = ±√((1 - cos(θ))/2) step 2 step 4 step 6 step 8

Answer

Explanation:

Step1: Analyze step 1

The double - angle formula $\cos(2x)=1 - 2\sin^{2}(x)$ is correct.

Step2: Analyze step 2

Letting $2x=\Theta$ is a valid substitution.

Step3: Analyze step 3

If $2x = \Theta$, then $x=\frac{\Theta}{2}$ is correct.

Step4: Analyze step 4

Substituting $x=\frac{\Theta}{2}$ into $\cos(2x)=1 - 2\sin^{2}(x)$ gives $\cos(\Theta)=1 - 2\sin^{2}(\frac{\Theta}{2})$, which is correct.

Step5: Analyze step 5

Starting from $\cos(\Theta)=1 - 2\sin^{2}(\frac{\Theta}{2})$, subtracting 1 from both sides gives $- 1+\cos(\Theta)=-2\sin^{2}(\frac{\Theta}{2})$, which is correct.

Step6: Analyze step 6

From $-1+\cos(\Theta)=-2\sin^{2}(\frac{\Theta}{2})$, we should get $1 - \cos(\Theta)=2\sin^{2}(\frac{\Theta}{2})$, not $1+\cos(\Theta)=2\sin^{2}(\frac{\Theta}{2})$. So step 6 has an error.

Step7: Analyze step 7

If it was the correct $1 - \cos(\Theta)=2\sin^{2}(\frac{\Theta}{2})$, then $\frac{1 - \cos(\Theta)}{2}=\sin^{2}(\frac{\Theta}{2})$ would be correct.

Step8: Analyze step 8

Taking the square - root of $\frac{1 - \cos(\Theta)}{2}=\sin^{2}(\frac{\Theta}{2})$ gives $\sin(\frac{\Theta}{2})=\pm\sqrt{\frac{1 - \cos(\Theta)}{2}}$, which is correct if step 6 was correct.

Answer:

step 6