Question Number 73757 by Maclaurin Stickker last updated on 15/Nov/19
$$\mathrm{cos}\:\mathrm{70}°+\mathrm{sin}\:\mathrm{200}°=? \\ $$
Commented by mathmax by abdo last updated on 15/Nov/19
$$\mathrm{70}^{{o}} =\frac{\mathrm{70}\pi}{\mathrm{180}}\:=\frac{\mathrm{7}\pi}{\mathrm{18}}\:\:\:{and}\:\mathrm{200}^{{o}} =\frac{\mathrm{200}\pi}{\mathrm{180}}\:=\frac{\mathrm{20}\pi}{\mathrm{18}}\:\Rightarrow \\ $$$${cos}\left(\mathrm{70}^{{o}} \right)+{sin}\left(\mathrm{200}^{{o}} \right)={cos}\left(\frac{\mathrm{7}\pi}{\mathrm{18}}\right)+{sin}\left(\frac{\mathrm{20}\pi}{\mathrm{18}}\right) \\ $$$$\frac{\mathrm{7}\pi}{\mathrm{18}}\:+\frac{\mathrm{20}\pi}{\mathrm{18}}\:=\frac{\mathrm{27}\pi}{\mathrm{18}}\:=\frac{\mathrm{3}\pi}{\mathrm{2}}\:\Rightarrow{sin}\left(\frac{\mathrm{20}\pi}{\mathrm{18}}\right)={sin}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}−\frac{\mathrm{7}\pi}{\mathrm{18}}\right) \\ $$$$={sin}\left(\pi\:+\frac{\pi}{\mathrm{2}}−\frac{\mathrm{7}\pi}{\mathrm{18}}\right)=−{sin}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{7}\pi}{\mathrm{18}}\right)=−{cos}\left(\frac{\mathrm{7}\pi}{\mathrm{18}}\right)\:\Rightarrow \\ $$$${cos}\left(\mathrm{70}^{{o}} \right)\:+{cos}\left(\mathrm{200}^{{o}} \right)=\mathrm{0} \\ $$
Answered by ajfour last updated on 15/Nov/19
$$=\mathrm{sin}\:\mathrm{20}°−\mathrm{sin}\:\mathrm{20}°=\mathrm{0} \\ $$
Commented by Maclaurin Stickker last updated on 15/Nov/19
$${Why}\:{it}\:{is}\:{sin}\:\mathrm{20}°? \\ $$
Commented by ajfour last updated on 15/Nov/19
$$\mathrm{cos}\:\left(\mathrm{70}°\right)=\mathrm{cos}\:\left(\mathrm{90}°−\mathrm{20}°\right)=\mathrm{sin}\:\mathrm{20}° \\ $$