trig 3

furor celtica · #1 $Old$ October 31st, 2009, 02:35 AM

express 4sinx + 3cosx in the form Rsin(x + A). hence find all the values of x in the range 0 to 360° for which cos3x=cos2x
so i found 4sinx + 3cosx= 5sin(x+36.9°), but after this i have no idea where to start to end up close to cos3x = cos2x. how do i connect these expressions?

Grandad · #2 $Old$ November 1st, 2009, 01:25 AM

Hello furor celtica

Quote:

Originally Posted by furor celtica $View Post$

express 4sinx + 3cosx in the form Rsin(x + A). hence find all the values of x in the range 0 to 360° for which cos3x=cos2x
so i found 4sinx + 3cosx= 5sin(x+36.9°), but after this i have no idea where to start to end up close to cos3x = cos2x. how do i connect these expressions?

I can't see any obvious way of connecting these two expressions either. All I can suggest is that you look at the identities

$\cos3x = 4\cos^3x-3\cos x$ and $\cos 2x = 2\cos^2x - 1$

together with

$\cos^2x = 1-\sin^2x$

and see if you can develop an expression like $4\sin x + 3\cos x$ .

But that seems an absurdly complicated method. The 'standard' way of solving an equation like $\cos3x=\cos2x$ is very simple. It uses the general solution of the equation $\cos A = \cos B$ , which is:

$A = 2n\pi \pm B, n = 0, \pm1, \pm2, ...$

and so you get:

$3x = 2n\pi \pm 2x$

$\Rightarrow x = 2n\pi$ or $5x = 2n\pi$

$\Rightarrow x = 2n\pi$ or $\tfrac25n\pi$

$\Rightarrow x = 0, \tfrac25\pi, \tfrac45\pi, \tfrac65\pi, \tfrac85\pi, 2\pi$ in the range $0\le x\le 2\pi$

Grandad

furor celtica · #3 $Old$ November 2nd, 2009, 12:27 AM

hm
i tried using those first identities and just lost a lot of time and scrap paper. are you sure there is no way to get to this using both relationships? this question was in the middle of other relatively simple ones, so i dont think that it would be overly complex