Why it does not work with a negative (trig problem)

[BOAS]

Hello,

i'm looking for an explanation of a rule I used to solve a problem. I came to the correct answer in the end, but I do not understand why my first attempt went wrong.

Homework Statement

Find the general solution to cot4θ = tan5θ

Homework Equations

tanA = tanB => A = nπ + B

tanα = a/b = cotβ

Where α and β are complementary angles.

The Attempt at a Solution

cot4θ = tan5θ

tan(π/2 - 4θ) = tan5θ

π/2 - 4θ = nπ + 5θ

9θ = π/2 - nπ

θ = π/18 (2n - 1)

The above is not given as a solution in my book and the following is said to be correct.

5θ = nπ + π/2 - 4θ

9θ = nπ + π/2

θ = π/18 (2n + 1)

Why is this?

Thanks!

 

[Saitama]

Hi BOAS!

Don't you think both the answers are same? (Check for different values of n) :)

 

[BOAS]

Hi BOAS!

Don't you think both the answers are same? (Check for different values of n) :)

I think for different values of n it is possible to get the same answer from both equations, but is that the same as 'being the same'?

IE - Is the first one not written in my book because it is the same as the second one?

 

[SammyS]

I think for different values of n it is possible to get the same answer from both equations, but is that the same as 'being the same'?

IE - Is the first one not written in my book because it is the same as the second one?

Both answers are the same.

θ = (π/18) multiplied by any odd integer .


Write your solution as
θ = (π/18)(2m-1)

The book solution is
θ = (π/18)(2n+1)

To get the same solution for both n = m-1 .