Question

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

Answer 1

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