Question

If x lies in the first quadrant and $\cos x=\frac{8}{17}$, then prove that:
$\cos \left(\frac{\mathrm{\pi }}{6}+x\right)+\cos \left(\frac{\mathrm{\pi }}{4}-x\right)+\cos \left(\frac{2\mathrm{\pi }}{3}-x\right)=\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right)\frac{23}{17}$

Answer 1

$$\begin{gathered} \text{Given}:0<x<\frac{\mathrm{\pi }}{2} \\ \mathrm{Now},\sin x=\sqrt{1-{\cos }^{2}x}=\sqrt{1-\frac{64}{289}}=\frac{15}{17} \\ \mathrm{LHS}=\cos \left(\frac{\mathrm{\pi }}{6}+x\right)+\cos \left(\frac{\mathrm{\pi }}{4}-x\right)+\cos \left(\frac{2\mathrm{\pi }}{3}-x\right) \\ =\cos (30+x)+\cos (45-x)+\cos (120-x) \\ =\cos 30°\cos x-\sin 30°\sin x+\cos 45°\cos x+\sin 45°\sin x+\cos 120°\cos x+\sin 120°\sin x\left\{\mathrm{Using}\mathrm{formulas}\mathrm{of}\cos (A+B)\mathrm{and}\cos (A-B\right\}) \\ =\cos x(\cos 30°+\cos 45°+\cos 120)+\sin x(-\sin 30°+\sin 45°+\sin 120°) \\ =\frac{8}{17}\left(\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2}\right)+\frac{15}{17}\left(-\frac{1}{2}+\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}\right) \\ =\frac{8}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right)+\frac{15}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right) \\ =\frac{23}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right) \\ =\mathrm{RHS} \\ \mathrm{Hence}\text{p}\mathrm{roved}. \end{gathered}$$