0* to 350*, find A
1. sin5A=sin3A
2.cos4A+cos2A=cos3A
Inverse Trigonometric Functions
Show, without using a calculator, that 〖Tan〗^(-1)4 -〖 Tan〗^(-1) 3/5 = π/4
General solution:
Case 1: $\displaystyle 5A = 3A + 2n \pi$ where n is an integer. Therefore $\displaystyle A = \, .... $
Case 2: $\displaystyle 5A = (\pi - 3A) + 2n\pi$ where n is an integer. Therefore $\displaystyle A = \, .... $
In each case substitute values of n that give values of A that lie in the given domain
You know that $\displaystyle \cos (x + y) + \cos (x - y) = 2 \cos x \cos y$. Substitute x = 3A and y = A: $\displaystyle \cos (4A) + \cos (2A) = 2 \cos (3A) \cos (A)$. Therefore the equation can be re-written as:
$\displaystyle 2 \cos (3A) \cos (A) = \cos (3A) \Rightarrow \cos (3A) ( 2 \cos (A) - 1) = 0$.
I'm sure you can continue the solution from here.
Let $\displaystyle \tan^{-1} (4) = \alpha \Rightarrow \tan \alpha = 4$.
Let $\displaystyle \tan^{-1} \left( \frac{3}{5} \right) = \beta \Rightarrow \tan \beta = \frac{3}{5}$.
Then you have to show that $\displaystyle \alpha - \beta = \frac{\pi}{4}$.
Now note that $\displaystyle \tan (\alpha -\beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = \frac{4 - \frac{3}{5}}{1 + (4)\left( \frac{3}{5}\right)} = \frac{20 - 3}{5 + 12} = 1$.
Therefore $\displaystyle \alpha - \beta = \tan^{-1} (1)$.
My mistake. I was using radians instead of degrees.
Case 1: It should be obvious that if $\displaystyle \sin (5A) = \sin (3A)$ then either 5A = 3A or the angles 5A and 3A differ by an integer multiple of $\displaystyle 360^0$.
So $\displaystyle 5A = 3A + 360^0 n$ where n is an integer.
Case 2: You should know that $\displaystyle \sin x = \sin (180^0 - x)$. Therefore $\displaystyle \sin (3A) = \sin (180^0 - 3A)$.
So the equation can be written $\displaystyle \sin (5A) = \sin (180^0 - 3A)$.
Using the same logic that gave Case 1, it should be obvious that either $\displaystyle 5A = (180^0 - 3A)$ or the angles $\displaystyle 5A$ and $\displaystyle (180^0 - 3A)$ differ by an integer multiple of $\displaystyle 360^0$.
So $\displaystyle 5A = (180^0 - 3A) + 360^0 n$ where n is an integer.
In each case, substitute the values of n into the general solutions that give values of A lying in the domain $\displaystyle 0^0 \leq A \leq 350^0$.