Atrisinājums

Pārveidojam doto sakarību:

\[\begin{aligned} & \frac{\cos 3x}{\cos x}=\frac{\cos (2x+x)}{\cos x}=\frac{\cos 2x \cos x-\sin 2x \sin x}{\cos x}=\frac{\cos 2x \cos x-2 \sin ^{2} x \cos x}{\cos x}= \\ & =\cos 2x-2 \sin ^{2} x=1-2 \sin ^{2} x-2 \sin ^{2} x=1-4 \sin ^{2} x=\frac{1}{2015} \end{aligned}\]

Izsakot, iegūstam \(4 \sin ^{2} x=\frac{2014}{2015}\). Aprēķināsim \(\frac{\sin 3x}{\sin x}\) vētību: \(\frac{\sin 3x}{\sin x}=\frac{\sin (2x+x)}{\sin x}=\frac{\sin 2x \cos x+\cos 2x \sin x}{\sin x}=\frac{2 \sin x \cos ^{2} x+\cos 2x \sin x}{\sin x}=\) \(=2 \cos ^{2} x+\cos 2x=2\left(1-\sin ^{2} x\right)+\left(1-2 \sin ^{2} x\right)=3-4 \sin ^{2} x=3-\frac{2014}{2015}=\frac{4031}{2015}\).