see this graph, only 6 solutions. figure out why?
I copied down the solutions from my teacher for this question and i understand up to this point:
⇒ 0 = sin 6x - sin 2x
The next step it is:
⇒ 0 = 2cos ((6x+2x)/2) sin ((6x-2x)/2)
I don't understand how that the addition formulas were used to get to this step.
The solution i copied down is below:
cos x + cos 3x = sin x + sin 3x
⇒cos x - sin x = sin 3x - cos 3x
now, squaring on both sides,
(cos x - sin x)^2 = (sin 3x - cos 3x)^2
⇒(cos x)^2 – 2 sin x cos x + (sin x)^2 = (sin 3x)^2 - 2sin 3x cos 3x+ (cos 3x)^2
⇒ 1-2 sin x cos x = 1-2sin 3x cos 3x
⇒ -2 sin x cos x = -2sin 3x cos 3x
⇒ sin x cos x = sin 3x cos 3x
multiplying both sides by 2,
⇒ 2sin x cos x = 2sin 3x cos 3x
⇒ sin 2x = sin 2(3x)
⇒ sin 2x = sin 6x
⇒ 0 = sin 6x - sin 2x
⇒ 0 = 2cos ((6x+2x)/2) sin ((6x-2x)/2)
⇒ 0 = 2cos 4x sin 2x
⇒ 0 = cos 4x sin 2x
case 1:
⇒ cos 4x = 0
⇒ cos 4x = cos π/2
⇒ 4x = 2nπ + π/2
⇒ x = nπ/2 + π/8
case 2:
⇒ sin 2x = 0
⇒ sin 2x = sin π
⇒ 2x = nπ + ((-1)^n)(π)
⇒ x = nπ/2 + ((-1)^n)(π)/2
⇒ x = [nπ/2 + π/8] or [nπ/2 + ((-1)^n)(π)/2]
Hello, BuffaloSoulja!
There are four sum-to-product identities:
. .
Your teacher's solution is rather silly . . .
He/she used (2) in that mysterious step.
What makes it silly is that he/she could have used these identities first
. . and save a lot of work.
We have: .
Using (1) and (3), we have: .
Factor: .
And we have two equations to solve:
. .