Thread: Intercepts for two trig equations

So I need a method to find the intercepts for two trig equations: y = sinx - cosx and y = 2cos(2x). I know that substitution is supposedly the easiest, but I can't seem to get anything usable. The segment I'm using is below. The actual intercepts are -1.1468, 1.3229 for the left and 0.7854, 0 for the right, but I need an algebraic/trigonometric/whatever way to do it.
Cheers.

Originally Posted by zapparage

So I need a method to find the intercepts for two trig equations: y = sinx - cosx and y = 2cos(2x). I know that substitution is supposedly the easiest, but I can't seem to get anything usable. The segment I'm using is below. The actual intercepts are -1.1468, 1.3229 for the left and 0.7854, 0 for the right, but I need an algebraic/trigonometric/whatever way to do it.
Cheers.

Solve sin x - cos x = 2 cos(2x):

sin x - cos x = 2 (cos^2 x - sin^2 x)

=> sin x - cos x = 2 (cos x - sin x)(cos x + sin x)

=> sin x - cos x = 0 or 1 = -2 (cos x + sin x)

=> tan x = 1 or cos x + sin x = -1/2.

To solve the second equation, note that cos x + sin x = sqrt{2} cos (x + pi/4).

Thanks a lot for the help, but I'm completely lost.

How'd you get:

Code:

 sin x - cos x = 2 (cos^2 x - sin^2 x)

and the 0 and 1 in:

Code:

=> sin x - cos x = 0 or 1 = 2 (cos x + sin x)

and also get to this:

Code:

=> tan x = 1 or cos x + sin x = 1/2.

Also, how do I get the exact intercept numbers once this is solved?

You've probably guessed that I have no idea about the 'sqrt' - we haven't even gotten that far.

Originally Posted by zapparage

Thanks a lot for the help, but I'm completely lost.

How'd you get:

Code:

 sin x - cos x = 2 (cos^2 x - sin^2 x)

Mr F asks: Have you been taught the double angle formulae?

and the 0 and 1 in:

Code:

=> sin x - cos x = 0 or 1 = 2 (cos x + sin x)

Mr F says: Factorise and use the null factor law:

sin x - cos x = 2 (cos x - sin x)(cos x + sin x) = -2 (sin x - cos x)(cos x + sin x)

=> 2 (sin x - cos x)(cos x + sin x) + (sin x - cos x) = 0

Common factor is sin x - cos x:

=> (sin x - cos x)(2(cos x + sin x) + 1) = 0

sin x - cos x = 0 or 2(cos x + sin x) + 1 = 0

etc.

and also get to this:

Code:

=> tan x = 1 or cos x + sin x = 1/2.

Mr F says: sin x - cos x = 0 => sin x = cos x => sin x/cos x = 1.

2(cos x + sin x) + 1 = 0 => 2(cos x + sin x) = -1 => cos x + sin x = -1/2.

Also, how do I get the exact intercept numbers once this is solved?

Mr F says: You solve the two trig equations to get the x-coordinates. Then you substitute the x-coordinates into either one of y = sinx - cosx and y = 2cos(2x) to get the y-coordinates.

You've probably guessed that I have no idea about the 'sqrt' - we haven't even gotten that far.

If you're required to get exact solutions for the x-intercepts, then it's expected that you:

1. Are familiar with the double angle formula for cos(2x).
2. Can re-arrange trig equations such as sin x - cos x = 0.
3. Can solve simple trig equations.
4. Can express cos x + sin x in the form I gave.