Proving trigonometric equation?

[skateza]

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

is this a proving trigonometric equation? i don't think it is

 

[D H]

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.

 

[skateza]

so how do u do that algebraically

 

[D H]

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).

 

[skateza]

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

i don't 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

 

[D H]

Start with cos(4x) = cos(2x+2x). Can you proceed from here?

 

[skateza]

ok i think that helped me realize the identity..

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

right?

 

[D H]

ok i think that helped me realize the identity..

Cos(A+B)=CosACosB-sinAsinB
cos(2x+2x)=cos2xcos2x-sin2xsin2x

Good so far.

=2cos2x-2sin2x

You are saying the equivalent of u*u = 2*u[/tex], which is obviously incorrect.

 

[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

 

[D H]

Use the identity \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).

 

[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.

 

[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)