If cos x = 8/17, Prove that cos(π/6+x) + cos(π/4−x) + cos(2π/3−x) = (23/17)[(√3−1)/2 + 1/√2]

Question

If

\[ \cos x=\frac{8}{17} \]

and \(x\) lies in the first quadrant, prove that:

\[ \cos\left(\frac{\pi}{6}+x\right) + \cos\left(\frac{\pi}{4}-x\right) + \cos\left(\frac{2\pi}{3}-x\right) \]

\[ = \frac{23}{17} \left( \frac{\sqrt3-1}{2} + \frac{1}{\sqrt2} \right) \]

Proof

Since

\[ \cos x=\frac{8}{17} \]

\[ \sin x=\sqrt{1-\cos^2x} \]

\[ = \sqrt{1-\frac{64}{289}} \]

\[ = \sqrt{\frac{225}{289}} = \frac{15}{17} \]

Now,

\[ \cos\left(\frac{\pi}{6}+x\right) = \cos\frac{\pi}{6}\cos x-\sin\frac{\pi}{6}\sin x \]

\[ = \frac{\sqrt3}{2}\cdot\frac{8}{17} – \frac12\cdot\frac{15}{17} \]

\[ = \frac{8\sqrt3-15}{34} \]

Also,

\[ \cos\left(\frac{\pi}{4}-x\right) = \cos\frac{\pi}{4}\cos x+\sin\frac{\pi}{4}\sin x \]

\[ = \frac{1}{\sqrt2}\cdot\frac{8}{17} + \frac{1}{\sqrt2}\cdot\frac{15}{17} \]

\[ = \frac{23}{17\sqrt2} \]

Also,

\[ \cos\left(\frac{2\pi}{3}-x\right) = \cos\frac{2\pi}{3}\cos x+\sin\frac{2\pi}{3}\sin x \]

\[ = -\frac12\cdot\frac{8}{17} + \frac{\sqrt3}{2}\cdot\frac{15}{17} \]

\[ = \frac{15\sqrt3-8}{34} \]

Adding,

\[ = \frac{8\sqrt3-15}{34} + \frac{23}{17\sqrt2} + \frac{15\sqrt3-8}{34} \]

\[ = \frac{23\sqrt3-23}{34} + \frac{23}{17\sqrt2} \]

\[ = \frac{23}{17} \left( \frac{\sqrt3-1}{2} + \frac{1}{\sqrt2} \right) \]

Hence proved.