Question

which expression is equivalent to ( 2cos^2left(\frac{x}{2}
ight) - cos(x) )? (\bigcirc) 0 (\bigcirc) 1 (\bigcirc) ( 1 - 2cos(x) ) (\bigcirc) ( 1 + 2cos(x) )

Explanation:

Step1: Recall the double - angle formula

We know the double - angle formula for cosine: \(\cos(2\theta)=2\cos^{2}\theta - 1\). We can rearrange this formula to solve for \(2\cos^{2}\theta\).
From \(\cos(2\theta)=2\cos^{2}\theta - 1\), we get \(2\cos^{2}\theta=1 + \cos(2\theta)\).

Step2: Substitute \(\theta=\frac{x}{2}\) into the formula

Let \(\theta = \frac{x}{2}\). Then \(2\cos^{2}(\frac{x}{2})=1+\cos(2\times\frac{x}{2})\).
Simplify the right - hand side: \(2\cos^{2}(\frac{x}{2})=1+\cos(x)\).

Step3: Substitute into the original expression

The original expression is \(2\cos^{2}(\frac{x}{2})-\cos(x)\). Substitute \(2\cos^{2}(\frac{x}{2}) = 1+\cos(x)\) into it:
\(2\cos^{2}(\frac{x}{2})-\cos(x)=(1 + \cos(x))-\cos(x)\).

Step4: Simplify the expression

Simplify \((1+\cos(x))-\cos(x)\):
\((1+\cos(x))-\cos(x)=1+\cos(x)-\cos(x)=1\).

Answer:

\(1 - \cos(x)\) (Wait, there is a mistake above. Let's re - solve it properly. The correct equivalent expression is \(1-\cos(x)\)? No, wait, let's use the double - angle formula. The double - angle formula for cosine is \(\cos(2\theta)=2\cos^{2}\theta - 1\), so \(2\cos^{2}\theta=1 + \cos(2\theta)\). Let \(\theta=\frac{x}{2}\), then \(2\cos^{2}(\frac{x}{2})=1+\cos(x)\). So \(2\cos^{2}(\frac{x}{2})-\cos(x)=1 + \cos(x)-\cos(x)=1\)? Wait, no, wait:

Wait, the double - angle formula: \(\cos(2\alpha)=2\cos^{2}\alpha - 1\), so \(2\cos^{2}\alpha=1+\cos(2\alpha)\). If \(\alpha=\frac{x}{2}\), then \(2\cos^{2}(\frac{x}{2})=1 + \cos(x)\). Then \(2\cos^{2}(\frac{x}{2})-\cos(x)=1+\cos(x)-\cos(x)=1\). So the answer is 1.