1. ## trigo >find angles

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

2. Originally Posted by sanikui
0* to 350*, find A
1. sin5A=sin3A
[snip]
General solution:

Case 1: $5A = 3A + 2n \pi$ where n is an integer. Therefore $A = \, ....$

Case 2: $5A = (\pi - 3A) + 2n\pi$ where n is an integer. Therefore $A = \, ....$

In each case substitute values of n that give values of A that lie in the given domain

3. Originally Posted by sanikui
0* to 350*, find A
[snip]
2.cos4A+cos2A=cos3A

[snip]
You know that $\cos (x + y) + \cos (x - y) = 2 \cos x \cos y$. Substitute x = 3A and y = A: $\cos (4A) + \cos (2A) = 2 \cos (3A) \cos (A)$. Therefore the equation can be re-written as:

$2 \cos (3A) \cos (A) = \cos (3A) \Rightarrow \cos (3A) ( 2 \cos (A) - 1) = 0$.

I'm sure you can continue the solution from here.

4. Originally Posted by sanikui
[snip]
Inverse Trigonometric Functions
Show, without using a calculator, that 〖Tan〗^(-1)4 -〖 Tan〗^(-1) 3/5 = π/4
Let $\tan^{-1} (4) = \alpha \Rightarrow \tan \alpha = 4$.

Let $\tan^{-1} \left( \frac{3}{5} \right) = \beta \Rightarrow \tan \beta = \frac{3}{5}$.

Then you have to show that $\alpha - \beta = \frac{\pi}{4}$.

Now note that $\tan (\alpha -\beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = \frac{4 - \frac{3}{5}}{1 + (4)\left( \frac{3}{5}\right)} = \frac{20 - 3}{5 + 12} = 1$.

Therefore $\alpha - \beta = \tan^{-1} (1)$.

5. Originally Posted by mr fantastic
General solution:

Case 1: $5A = 3A + 2n \pi$ where n is an integer. Therefore $A = \, ....$

Case 2: $5A = (\pi - 3A) + 2n\pi$ where n is an integer. Therefore $A = \, ....$

In each case substitute values of n that give values of A that lie in the given domain
This i till cant understand...can explain more clearly?
Next 2 question i can get ur mean.Thx.

6. Originally Posted by mr fantastic
General solution:

Case 1: $5A = 3A + 2n \pi$ where n is an integer. Therefore $A = \, ....$

Case 2: $5A = (\pi - 3A) + 2n\pi$ where n is an integer. Therefore $A = \, ....$

In each case substitute values of n that give values of A that lie in the given domain

Case 1: It should be obvious that if $\sin (5A) = \sin (3A)$ then either 5A = 3A or the angles 5A and 3A differ by an integer multiple of $360^0$.

So $5A = 3A + 360^0 n$ where n is an integer.

Case 2: You should know that $\sin x = \sin (180^0 - x)$. Therefore $\sin (3A) = \sin (180^0 - 3A)$.

So the equation can be written $\sin (5A) = \sin (180^0 - 3A)$.

Using the same logic that gave Case 1, it should be obvious that either $5A = (180^0 - 3A)$ or the angles $5A$ and $(180^0 - 3A)$ differ by an integer multiple of $360^0$.

So $5A = (180^0 - 3A) + 360^0 n$ 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 $0^0 \leq A \leq 350^0$.

7. Oic...know liaw....thx Mr.fantastic~!!