Evaluating cos((1/2)arccos(x))

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SUMMARY

The discussion focuses on evaluating the expression cos((1/2)arccos(x)). It establishes that the range of arccos(x) is between 0 and π, leading to the conclusion that cos((1/2)arccos(x)) will yield non-negative results. The half-angle identity for cosine is introduced, specifically cos²(θ/2) = (1 + cos(θ))/2, allowing for the simplification of the expression to cos(θ/2) = √((1 + cos(θ))/2). This provides a clear pathway for further evaluation of the expression.

PREREQUISITES
  • Understanding of trigonometric functions and their inverses, specifically arccos(x).
  • Familiarity with the half-angle identities in trigonometry.
  • Basic knowledge of the properties of the cosine function.
  • Ability to manipulate algebraic expressions involving square roots.
NEXT STEPS
  • Study the half-angle identities in trigonometry for deeper insights.
  • Practice problems involving the evaluation of arccos(x) and its applications.
  • Explore the properties of the cosine function and its range.
  • Learn about the implications of non-negative outputs in trigonometric evaluations.
USEFUL FOR

Students studying trigonometry, educators teaching trigonometric identities, and anyone needing assistance with evaluating expressions involving inverse trigonometric functions.

Elissa89
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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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