Question

Prove that $\sqrt{2+\sqrt{2+\sqrt{2+2cos8\theta }}}=2cos\theta .$

Or

Prove that $(1+cos\frac{\pi }{8})(1+cos\frac{3\pi }{8})(1+cos\frac{5\pi }{8})(1+cos\frac{7\pi }{8})=\frac{1}{8}$

Answer 1

LHS = $\sqrt{2+\sqrt{2+\sqrt{2+2cos8\theta }}}$

= $\sqrt{2+\sqrt{2+\sqrt{2(1+cos8\theta )}}}$

= $\sqrt{2+\sqrt{2+\sqrt{2.2co{s}^{2}4\theta }}}[\because 2co{s}^2\theta =1+cos2\theta ]$

= $\sqrt{2+\sqrt{2+2cos4\theta }}=\sqrt{2+\sqrt{2(1+cos4\theta )}}$

= $\sqrt{2+\sqrt{2.2co{s}^{2}2\theta }}[\because 2co{s}^2\theta =1+cos2\theta ]$

= $\sqrt{2+2cos2\theta }$

= $\sqrt{2(1+cos2\theta )}$

= $\sqrt{2.2co{s}^2\theta }=2cos\theta =RHS$ Hence proved.

Or

LHS = $(1+cos\frac{\pi }{8})(1+cos\frac{3\pi }{8})(1+cos\frac{5\pi }{8})(1+cos\frac{7\pi }{8})$

= $(1+cos\frac{\pi }{8})(1+cos\frac{3\pi }{8})[1+cos(\pi -\frac{3\pi }{8})][1+cos(\pi -\frac{\pi }{8})]$

= $(1+cos\frac{\pi }{8})(1+cos\frac{3\pi }{8})(1-cosw\frac{3\pi }{8})(1-cos\frac{\pi }{8})$

= $(1+cos\frac{\pi }{8})(1-cos\frac{\pi }{8})(1+cos\frac{3\pi }{8})(1-cos\frac{3\pi }{8})$

= $(1-co{s}^2\frac{\pi }{8})(1-co{s}^2\frac{3\pi }{8})=si{n}^2\frac{\pi }{8}\times si{n}^2\frac{3\pi }{8}$

$[\because 1-co{s}^2\theta =si{n}^2\theta ]$

= $\left(\frac{1-cos\pi /4}{2}\right)\left(\frac{1-cos\frac{3\pi }{4}}{2}\right)$

$[\because si{n}^{2}x=\frac{1-cos2x}{2}]$

= $\frac{1}{4}(1-\frac{1}{\sqrt{2}})(1+\frac{1}{\sqrt{2}})$

= $\frac{1}{4}(1-\frac{1}{2})=\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}=RHS$ Hence proved.