Question

Rewrite with only sin x and cos x.

sin 3x - cos

Answer 1

Use the sum & difference formulas.

sin3x+cos3x = sin (x + 2x) + cos (x + 2x)
= (sin x)(cos 2x) + (cos x)(sin 2x) + (cos x)(cos 2x) - (sin x)(sin 2x)

Now sin 2x = 2sin x cos x, and cos 2x = (cos x)^2 - (sin x)^2
=(sin x)((cos x)^2 - (sin x)^2) + (cos x)(2(sin x)(cos x)) + (cos x)((cos x)^2 - (sin x)^2) - (sin x)(2(sin x)(cos x))
= (sin x)(cos x)^2 - (sin x)^3 + 2(sin x)(cos x)^2 + (cos x)^3 - (cos x)(sin x)^2 - 2(cos x)(sin x)^2

= (cos x)^3 + 3(sin x)(cos x)^2 - 3(cos x)(sin x)^2 - (sin x)^3

Answer 2

I don't understand

Answer 3

The options are:
2 sin x - sin3x - cos x

2 sin x cos2x + sin x - 2 sin3x - cos x

2 sin3x cos4x + 1

3 sin x cos x - sin3x - cos x

Answer 4

2 sin x - sin^3x - cos x

2 sin x cos^2x + sin x - 2 sin^3x - cos x

2 sin^3x cos^4x + 1

3 sin x cos x - sin^3x - cos x

Answer 5

@april

Answer 6

@i0iz0gangster