Question

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

Answer 1

We have,
$cos\theta =\frac{8}{17}$
$\therefore sin\theta =\sqrt{1-co{s}^2\theta }=\sqrt{1-\frac{64}{289}}$
$=\sqrt{\frac{225}{289}}$
$=\frac{15}{17}$
Now, $cos(\frac{\pi }{6}+\theta )+cos(\frac{\pi }{4}-\theta )+cos(\frac{2\pi }{3}-\theta )$
$=[cos\frac{\pi }{6}cos\theta -sin\frac{\pi }{6}sin\theta ]+[cos\frac{\pi }{4}cos\theta +sin\frac{\pi }{4}sin\theta ]+[cos\frac{2\pi }{3}cos\theta +sin\frac{2\pi }{3}sin\theta ]$
$=[cos\frac{\pi }{6}+cos\frac{\pi }{4}+cos\frac{2\pi }{3}]cos\theta +sin\theta [-sin\frac{\pi }{6}+sin\frac{\pi }{4}+sin\frac{2\pi }{3}]$
$=[\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}+cos(\frac{\pi }{2}+\frac{\pi }{6})]\times \frac{8}{17}+\frac{15}{17}\times [-\frac{1}{2}+\frac{1}{\sqrt{2}}+sin(\frac{\pi }{2}+\frac{\pi }{6})]$
$=[\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-sin\frac{\pi }{6}]\times \frac{8}{17}+\frac{15}{17}\times [-\frac{1}{2}+\frac{1}{\sqrt{2}}+cos\frac{\pi }{6}]$
$=[\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2}]\times \frac{8}{17}+\frac{15}{17}\times [-\frac{1}{2}+\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}]$
$=[\frac{\sqrt{3-1}}{2}+\frac{1}{\sqrt{2}}]\times \frac{8}{17}+\frac{15}{17}\times [\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}]$
$=(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}})(\frac{8}{17}+\frac{15}{17})$
$=(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}})\left(\frac{8+15}{17}\right)$
$=(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}})\times \frac{23}{17}$
$\therefore cos(\frac{\pi }{6}+\theta )+cos(\frac{\pi }{4}-\theta )+cos(\frac{2\pi }{3}-\theta )=(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}})\times \frac{23}{17}$
Hence proved.