Solve. 1. sin x cos^3 x-cosx sin^3 x=1/4sin^4x (cos x+cos 4x+cos 7x)/(sin x+sin 4x+sin 7x)=cot 4x

Question

Solve.
1.

\sin x \cos^{3} x - \cos x \sin^{3} x = 1 / 4 \sin 4 x

2.

\frac{\cos x + \cos 4 x + \cos 7 x}{\sin x + \sin 4 x + \sin 7 x} = \cot 4 x

Answer 1

1) Take out common factors:

\sin x \cos x [\cos^{2} x - \sin^{2} x] = (1 / 4) \sin (4 x)

double angle identity:

\sin (2 x) = 2 \sin (x) \cos (x)

So,

\frac{\sin (2 x)}{2} [\cos (2 x)] = \frac{\sin (4 x)}{4}

\frac{\sin (2 x) \cos (2 x)}{2} = \frac{\sin (4 x)}{4}

We already know that:

\sin (2 u) = \sin (u) \cos (u)

if we let u = 2x:

\sin [2 (2 x)] = \sin (2 x) \cos (2 x)

\sin (4 x) = \sin (2 x) \cos (2 x)

\frac{\sin (4 x)}{4} = \frac{\sin (4 x)}{4}

2) we know that

\cos x + \cos 7 x = 2 \cos (7 x + x) / 2 * c o s (7 x - x) / 2 = 2 \cos 4 x \cos 3 x

and

\sin x + \sin 7 x = 2 \sin (7 x + x) / 2 * \cos (7 x - x) / 2 = 2 \sin 4 x \cos 3 x

\frac{\cos x + \cos 4 x + \cos 7 x}{\sin x + \sin 4 x + \sin 7 x} = \frac{2 \cos 4 x * \cos 3 x + \cos 4 x}{2 \sin 4 x * \cos 3 x + \sin 4 x} = \frac{\cos 4 x (2 \cos 3 x + 1)}{\sin 4 x (2 \cos 3 x + 1)} = \frac{\cos 4 x}{\sin 4 x} = \cot x

Answers (1)