# How do you prove [(sin 3x) / sin x ] – [(cos3x) / cos x)] = 2?

$\frac{\sin 3 x}{\sin} x - \frac{\cos 3 x}{\cos} x = 2 \text{ }$TRUE, it is an identity

#### Explanation:

$\frac{\sin 3 x}{\sin} x - \frac{\cos 3 x}{\cos} x = 2$

$\frac{\sin \left(2 x + x\right)}{\sin} x - \frac{\cos \left(2 x + x\right)}{\cos} x = 2$

$\frac{\sin 2 x \cdot \cos x + \cos 2 x \cdot \sin x}{\sin} x - \frac{\cos 2 x \cdot \cos x - \sin 2 x \cdot \sin x}{\cos} x = 2$

$\frac{\sin 2 x \cdot \cos x}{\sin} x + \frac{\cos 2 x \cdot \sin x}{\sin} x - \frac{\cos 2 x \cdot \cos x}{\cos} x + \frac{\sin 2 x \cdot \sin x}{\cos} x = 2$

$\frac{\sin 2 x \cdot \cos x}{\sin} x + \frac{\cos 2 x \cdot \cancel{\sin} x}{\cancel{\sin}} x - \frac{\cos 2 x \cdot \cancel{\cos} x}{\cancel{\cos}} x + \frac{\sin 2 x \cdot \sin x}{\cos} x = 2$

$\frac{\sin 2 x \cdot \cos x}{\sin} x + \cos 2 x - \cos 2 x + \frac{\sin 2 x \cdot \sin x}{\cos} x = 2$

$\frac{\sin 2 x \cdot \cos x}{\sin} x + \frac{\sin 2 x \cdot \sin x}{\cos} x = 2$

$\frac{2 \cdot \sin x \cdot \cos x \cdot \cos x}{\sin} x + \frac{2 \cdot \sin x \cdot \cos x \cdot \sin x}{\cos} x = 2$

$\frac{2 \cdot \cancel{\sin} x \cdot \cos x \cdot \cos x}{\cancel{\sin}} x + \frac{2 \cdot \sin x \cdot \cancel{\cos} x \cdot \sin x}{\cancel{\cos}} x = 2$

$\left(2 \cdot \cos x \cdot \cos x\right) + \left(2 \cdot \sin x \cdot \sin x\right) = 2$

$2 \cdot {\cos}^{2} x + 2 \cdot {\sin}^{2} x = 2$

$2 \cdot \left({\cos}^{2} x + {\sin}^{2} x\right) = 2$

$2 \cdot \left(1\right) = 2$

$2 = 2 \text{ }$it is an IDENTITY

God bless....I hope the explanation is useful.

Jul 1, 2016

$L . H . S . = \frac{\sin 3 x \cos x - \cos 3 x \sin x}{\sin x \cos x}$

Recall here that $\sin A \cos B - \cos A \sin B = \sin \left(A - B\right)$, Later we will also use $\sin 2 x = 2 \sin x \cos x$.

$\therefore L . H . S . = \sin \frac{3 x - x}{\sin x \cos x} = \frac{\sin 2 x}{\sin x \cos x} = \frac{2 \sin x \cos x}{\sin x \cos x} = 2 = R . H . S .$

Yet another way to prove this Identity [as, Mr. Leland Adriano Alejandro has rightly said it!] is to use the Multiple Angle Formula for $\cos 3 x = 4 {\cos}^{3} x - 3 \cos x$, and, $\sin 3 x = 3 \sin x - 4 {\sin}^{3} x$.

$\therefore L . H . S . = \frac{3 \sin x - 4 {\sin}^{3} x}{\sin} x - \frac{4 {\cos}^{3} x - 3 \cos x}{\cos} x$
$= \frac{\sin x \left(3 - 4 {\sin}^{2} x\right)}{\sin} x - \frac{\cos x \left(4 {\cos}^{2} x - 3\right)}{\cos} x$
$= 3 - 4 {\sin}^{2} x - 4 {\cos}^{2} x + 3$
$= 6 - 4 \left({\sin}^{2} x + {\cos}^{2} x\right) = 6 - 4 = 2$, as proved before!

Hope, you will enjoy the proofs!

Jul 1, 2016

As follows

#### Explanation:

To prove

$\frac{\sin 3 x}{\sin} x - \frac{\cos 3 x}{\cos} x = 2$

Method-1 (shortest)

$L H S = \frac{\sin 3 x}{\sin} x - \frac{\cos 3 x}{\cos} x$

=(sin3xcosx-cos3xsinx)/(sinxcosx

$= \sin \frac{3 x - x}{\sin x \cos x}$

$= \sin \frac{2 x}{\sin x \cos x} = \frac{2 \sin x \cos x}{\sin x \cos x} = 2$

Using identity $\sin 2 x = 2 \sin x \cos x$
proved

Method -2
Using identities

$\sin 3 x = \left(3 \sin x - 4 {\sin}^{3} x\right) \mathmr{and} \cos 3 x = \left(4 {\cos}^{3} x - 3 \cos x\right)$

$L H S = \frac{\sin 3 x}{\sin} x - \frac{\cos 3 x}{\cos} x$

$= \frac{3 \sin x - 4 {\sin}^{3} x}{\sin} x - \frac{4 {\cos}^{3} x - 3 \cos x}{\cos} x$

$= \frac{3 \sin x}{\sin} x - \frac{4 {\sin}^{3} x}{\sin} x - \frac{4 {\cos}^{3} x}{\cos} x + \frac{3 \cos x}{\cos} x$

$= 3 - 4 {\sin}^{2} x - 4 {\cos}^{2} x + 3$

$= 6 - 4 \left({\sin}^{2} x + {\cos}^{2} x\right) = 6 - 4 \times 1 = 2$

[ since $\left({\sin}^{2} x + {\cos}^{2} x\right) = 1$]

proved