MHB Evaluating cos((1/2)arccos(x))

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To evaluate cos((1/2)arccos(x)), it's important to recognize that 0 ≤ arccos(x) ≤ π, which implies 0 ≤ (1/2)arccos(x) ≤ (π/2), ensuring the cosine value is non-negative. The half-angle identity for cosine can be applied: cos²(θ/2) = (1 + cos(θ))/2. This leads to the conclusion that cos(θ/2) can be expressed as the square root of (1 + cos(θ))/2. The discussion emphasizes the need to apply these identities correctly to solve the problem.
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I don't know how to solve this, we didn't really cover any problems like this in class

cos(1/2*cos^-1*x)

This is due tonight online and would like help please.
 
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Re: Need help, due tonight

The first thing I would consider is:

$$0\le\arccos(x)\le\pi$$

Hence:

$$0\le\frac{1}{2}\arccos(x)\le\frac{\pi}{2}$$

This means the cosine of the given angle will be non-negative. Next, consider the half-angle identity for cosine:

$$\cos^2\left(\frac{\theta}{2}\right)=\frac{1+\cos(\theta)}{2}$$

Given that the cosine function will be non-negative, we may write:

$$\cos\left(\frac{\theta}{2}\right)=\sqrt{\frac{1+\cos(\theta)}{2}}$$

Can you proceed?
 
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