Question Number 17983 by alex041103 last updated on 13/Jul/17
$${Evaluate}\:{cos}\left(\mathrm{20}°\right){cos}\left(\mathrm{40}°\right){cos}\left(\mathrm{80}°\right). \\ $$$${This}\:{question}\:{is}\:{just}\:{for}\:{fun}\:{and}\:{practice}. \\ $$$${Evryone}\:{who}\:{wants}\:{can}\:{answer}\:{this}\:{question}. \\ $$
Answered by ajfour last updated on 13/Jul/17
![cos 20°cos 40°cos 80°= (1/(2sin 20°))[2sin 20°cos 20°]cos 40°cos 80° =(1/(4sin 20°))[2sin 40°cos 40°]cos 80° =(1/(8sin 20°))[2sin 80°cos 80°] = ((sin (180°−20°))/(8sin 20°)) =(1/8) .](https://www.tinkutara.com/question/Q17984.png)
$$\mathrm{cos}\:\mathrm{20}°\mathrm{cos}\:\mathrm{40}°\mathrm{cos}\:\mathrm{80}°= \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{2sin}\:\mathrm{20}°}\left[\mathrm{2sin}\:\mathrm{20}°\mathrm{cos}\:\mathrm{20}°\right]\mathrm{cos}\:\mathrm{40}°\mathrm{cos}\:\mathrm{80}° \\ $$$$=\frac{\mathrm{1}}{\mathrm{4sin}\:\mathrm{20}°}\left[\mathrm{2sin}\:\mathrm{40}°\mathrm{cos}\:\mathrm{40}°\right]\mathrm{cos}\:\mathrm{80}° \\ $$$$=\frac{\mathrm{1}}{\mathrm{8sin}\:\mathrm{20}°}\left[\mathrm{2sin}\:\mathrm{80}°\mathrm{cos}\:\mathrm{80}°\right] \\ $$$$=\:\frac{\mathrm{sin}\:\left(\mathrm{180}°−\mathrm{20}°\right)}{\mathrm{8sin}\:\mathrm{20}°}\:=\frac{\mathrm{1}}{\mathrm{8}}\:. \\ $$
Commented by alex041103 last updated on 13/Jul/17
$${well}\:{done}\:{sir} \\ $$
Commented by alex041103 last updated on 13/Jul/17
$${if}\:{you}\:{want}\:{you}\:{can}\:{try}\:{Q}.\mathrm{17525} \\ $$