trig 3

[red and white kop!]

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?

 

[chrisr]

belatedly,
maybe, the "hence" is an "indirect reference" to the technique of using identities.

Cos3x=Cos2x
Cos(A+B)=Cos(A-B)
A+B=3x and A-B=2x, so 2A=5x, A=5x/2 and B=x/2.

Cos(5x/2 + x/2) = Cos(5x/2 - x/2)
Cos(5x/2)Cos(x/2)-Sin(5x/2)Sinx/2 = Cos(5x/2)Cos(x/2)+Sin(5x/2)Sin(x/2)
2Sin(5x/2)Sin(x/2)=0, so Sin(5x/2)Sin(x/2)=0.

Sin(5x/2) =0 and Sin(x/2) =0 gives x=0, 2(pi)/5, 4(pi)/5, 6(pi)/5, 8(pi)/5, 2(pi).

Cos{6(pi)/5} = Cos{4(pi)/5} etc. as Cosx is the horizontal co-ordinate.

 

[fasteddie65]

cos 3x = cos 2x
cos (2x + x) = cos 2x
cos 2x cos x - sin 2x sin x = cos 2x
(2 cos^2 x - 1) cos x - (2 sin x cos x) sin x = cos 2x
(2 cos^2 x - 1) cos x - 2 sin^2 x cos x = cos 2x
2 cos^3 x - cos x - 2(1 - cos^2 x) cos x = cos 2x
2 cos^3 x - cos x - 2 cos x + 2 cos^3 x = 2 cos^2 x - 1
4 cos^3 x - 2 cos^2 x - 3 cos x + 1 = 0
cos x = -0.8091..., 0.3090..., 1

It's easy to show that x = 0; the other two values of x can be found using a calculator.