# Proving trigonometric equation?

1. Sep 20, 2007

### skateza

for x between 0 and 2pie, solve cos(4x)=sin(2x)....

is this a proving trigonometric equation? i dont think it is

2. Sep 20, 2007

### D H

Staff Emeritus
This is not a tautology, which you should be able to see if you expand cos(4x) in terms of sin(2x). The problem is find the value(s) of x between 0 and 2 pi for which the given expression is true.

3. Sep 20, 2007

### skateza

so how do u do that algebraically

4. Sep 20, 2007

### D H

Staff Emeritus
I gave you a big hint in my first post. I'll repeat to make it blatantly obvious: Expand cos(4x) in terms of sin(2x).

5. Sep 20, 2007

### skateza

i understand that,i'm not an idiot i'm just missing something very crucial to be able to determine the answer.

i dont know how to expand it into terms of sin, i have checked through my textbook i have looked everywhere. I know this is probably really easy but i'm missing that key concept which i can't figure out to be able to put cos in terms of sin

6. Sep 20, 2007

### D H

Staff Emeritus

7. Sep 20, 2007

### skateza

ok i think that helped me realise the identity..

Cos(A+B)=CosACosB-sinAsinB
cos(2x+2x)=cos2xcos2x-sin2xsin2x
=2cos2x-2sin2x
2cos2x-2sin2x=sin2x
2cos2x=3sin2x
2/3=tan2x?

right?

8. Sep 20, 2007

### D H

Staff Emeritus
Good so far.
You are saying the equivalent of $u*u = 2*u[/tex], which is obviously incorrect. 9. Sep 20, 2007 ### skateza Ok so this is what i have so far cos4x=sin2x cos(2x+2x)=sin2x cos^22x-sin^22x=sin2x as far as i know, cos^2x+sin^2x=1, so when i have -sin^22x, i can't complete that property correct? or is it just -1 10. Sep 20, 2007 ### D H Staff Emeritus Use the identity [itex]\cos^2(2x)+\sin^2(2x) = 1$ to eliminate the $\cos^2(2x)$ term: $\cos^2(2x)-\sin^2(2x) = 1 - 2\sin^2(2x)$. Applying this to the original problem yields

$$1 - 2\sin^2(2x) = \sin(2x)$$

which is a quadratic equation in $\sin(2x)$.

11. Sep 22, 2007

### Curious3141

A more elementary solution would be to utilise the identity

$$\sin(\frac{\pi}{2} \pm y) = \cos y$$

Since multiple angles are in play here, add 2n*pi to the argument, for integral n.

$$\sin(2n\pi + \frac{\pi}{2} \pm y) = \cos y$$

giving $$\sin(\frac{1}{2}(4n + 1)\pi \pm y) = \cos y$$

Now substitute that into the cosine expression in the LHS of the orig. equation (y = 4x), remove the sines on both sides, and you have a linear equation to solve. Simply list the multiple solutions in the required range by varying n (n can be zero, positive or negative). Don't have to worry about the plus/minus part too much, since all the solutions with one sign are included when solving for the other, but you need to establish this.

12. Sep 23, 2007

### silver-rose

Furthering D H's help,

After subbing cos (4x) for 1- sin^2 (2x), make a substitution to form a simple quadratic equation which you can solve (ie. let sin 2x = u)