Question

If ${\cos }^2\left(\frac{\mathrm{\pi }}{6}+x\right)-{\sin }^2\left(\frac{\mathrm{\pi }}{6}-x\right)=k\cos 2x$ then k = _____________.

Answer 1

$$\begin{gathered} \mathrm{Given} \\ {\cos }^2\left(\frac{\mathrm{\pi }}{6}+x\right)-{\sin }^2\left(\frac{\mathrm{\pi }}{6}-x\right)=K\cos 2x \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\mathrm{is}{\left(\cos \left(\frac{\mathrm{\pi }}{6}+x\right)\right)}^2-{\left(\sin \left(\frac{\mathrm{\pi }}{6}-x\right)\right)}^2 \\ ={\left(\cos \frac{\mathrm{\pi }}{6}\cos x-\sin \frac{\mathrm{\pi }}{6}\sin x\right)}^2-{\left(\sin \frac{\mathrm{\pi }}{6}\cos x-\cos \frac{\mathrm{\pi }}{6}\sin x\right)}^2 \\ \mathrm{using}\mathrm{identities}:- \\ \cos \left(a+b\right)=\cos a\cos b-\sin a\sin b \\ \mathrm{and}\sin \left(a-b\right)=\sin a\cos b-\cos a\sin b \\ ={\left(\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x\right)}^2-{\left(\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x\right)}^2 \\ =\frac{3}{4}{\cos }^{2}x+\frac{1}{4}{\sin }^{2}x-\frac{\sqrt{3}}{2}\sin x\cos x-\left[\frac{1}{4}{\cos }^{2}x+\frac{3}{4}{\sin }^{2}x-\frac{\sqrt{3}}{2}\sin x\cos x\right] \\ =\frac{3}{4}{\cos }^{2}x-\frac{1}{4}{\cos }^{2}x+\frac{1}{4}{\sin }^{2}x-\frac{3}{4}{\sin }^{2}x \\ =\frac{1}{2}{\cos }^{2}x-\frac{1}{2}{\sin }^{2}x \\ =\frac{1}{2}\left({\cos }^{2}x-{\sin }^{2}x\right) \\ =\frac{1}{2}\cos 2x=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{i}.\mathrm{e}.K\cos 2x \\ \Rightarrow K=\frac{1}{2} \\ \therefore \mathrm{value}\mathrm{of}K\mathrm{is}\frac{1}{2}. \end{gathered}$$