1-cos(5*a)+cos(5*b)=0 and 1-cos(7*a)+cos(7*b)=0  .solve for  a and b

We are given two trigonometric equations:

\[
1 - \cos(5a) + \cos(5b) = 0 \quad \text{(1)}
\]
\[
1 - \cos(7a) + \cos(7b) = 0 \quad \text{(2)}
\]

### Step 1: Rearrange each equation

#### Equation (1):
Rearranging for \(\cos(5a)\):

\[
1 - \cos(5a) + \cos(5b) = 0
\]
\[
\cos(5a) = 1 + \cos(5b) \quad \text{(3)}
\]

#### Equation (2):
Similarly, rearrange equation (2):

\[
1 - \cos(7a) + \cos(7b) = 0
\]
\[
\cos(7a) = 1 + \cos(7b) \quad \text{(4)}
\]

### Step 2: Apply cosine bounds

The cosine function has a range of \([-1, 1]\), meaning for any real value of \(a\) and \(b\), the values of \(\cos(5a)\) and \(\cos(7a)\) must lie within this range. This helps us consider special cases when analyzing further.

### Step 3: Try specific values

We can test specific values of \(a\) and \(b\) to see if they satisfy both equations simultaneously. A logical choice is \(a = b\), as it might lead to a simpler solution.

If \(a = b\), then from equation (3):

\[
\cos(5a) = 1 + \cos(5a)
\]

This simplifies to:

\[
\cos(5a) = 1
\]

Thus, \(5a = 2n\pi\), meaning:

\[
a = \frac{2n\pi}{5} \quad \text{for integer } n.
\]

Now apply the same logic to equation (4):

\[
\cos(7a) = 1 + \cos(7a)
\]

This simplifies to:

\[
\cos(7a) = 1
\]

Thus, \(7a = 2m\pi\), meaning:

\[
a = \frac{2m\pi}{7} \quad \text{for integer } m.
\]

### Step 4: Solve for \(a\) and \(b\)

For \(a = b\), both \(a\) and \(b\) must satisfy:

\[
\frac{2n\pi}{5} = \frac{2m\pi}{7}
\]

This implies \(5m = 7n\), so \(m = 7k\) and \(n = 5k\) for some integer \(k\). Therefore:

\[
a = b = \frac{2k\pi}{35}
\]

### Conclusion:

The general solutions for \(a\) and \(b\) are:

\[
a = b = \frac{2k\pi}{35} \quad \text{for integer } k.
\]

Would you like further details or clarifications on any step?

### Related Questions:
1. What is the solution if \(a \neq b\)?
2. How does the cosine function behave for large values of \(a\) and \(b\)?
3. Can you solve the equations graphically?
4. What happens if we use sine functions instead of cosine?
5. How does the solution change if \(a\) or \(b\) are complex?

**Tip:** Cosine values are periodic, so checking the behavior at specific intervals like multiples of \(\pi\) often simplifies the problem.

1 - cos(5*a) + cos(5*b) = 0 and 1 - cos(7*a) + cos(7*b) = 0 .solve for a and b

Discover the step-by-step process for solving trigonometric equations like 1 - cos(5a) + cos(5b) = 0 and 1 - cos(7a) + cos(7b) = 0. This advanced solution explains key concepts in trigonometry, including cosine function behavior, periodicity, and how to solve for variables a and b. Ideal for college-level math students.

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