Trigonometric Function prove Problem

**mastermin346** · October 23rd 2011, 10:28 PM

If $Sin 2\theta \ne 0$ prove that $\frac{sin3\theta}{sin\theta} - \frac{cos3\theta}{cos\theta} = 2$ .hence or otherwise,Show that $\frac{sin^{2}3\theta}{sin^{2}\theta} - \frac{cos^{2}3\theta}{cos^{2}\theta} = 8cos 2\theta$

**Prove It** · October 23rd 2011, 10:41 PM

Originally Posted by mastermin346

If $Sin 2\theta \ne 0$ prove that $\frac{sin3\theta}{sin\theta} - \frac{cos3\theta}{cos\theta} = 2$ .hence or otherwise,Show that $\frac{sin^{2}3\theta}{sin^{2}\theta} - \frac{cos^{2}3\theta}{cos^{2}\theta} = 8cos 2\theta$

You should know that $\displaystyle \sin{(\alpha \pm \beta)} \equiv \sin{(\alpha)}\cos{(\beta)} \pm \cos{(\alpha)}\sin{(\beta)}$ and $\displaystyle \cos{(\alpha \pm \beta)} \equiv \cos{(\alpha)}\cos{(\beta)} \mp \sin{(\alpha)}\sin{(\beta)}$

Here use $\displaystyle \sin{(3\theta)} = \sin{(2\theta + \theta)}$ and $\displaystyle \cos{(3\theta)} = \cos{(2\theta + \theta)}$

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