MHB Evaluating cos((1/2)arccos(x))

  • Thread starter Thread starter Elissa89
  • Start date Start date
Click For Summary
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.
Elissa89
Messages
52
Reaction score
0
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.
 
Last edited:
Mathematics news on Phys.org
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?
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K