Question

If x lies in the first quadrant and $$\cos x=\frac{8}{17}$$, then prove that:

$$\cos (\frac{\pi }{6}+x)+\cos (\frac{\pi }{4}-x)+\cos (\frac{2\pi }{3}-x)=(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}})\frac{23}{17}$$

 

Answer 1

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