cos-70-sin-200-

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

70^o =((70π)/(180)) =((7π)/(18)) and 200^o =((200π)/(180)) =((20π)/(18)) ⇒ cos(70^o )+sin(200^o )=cos(((7π)/(18)))+sin(((20π)/(18))) ((7π)/(18)) +((20π)/(18)) =((27π)/(18)) =((3π)/2) ⇒sin(((20π)/(18)))=sin(((3π)/2)−((7π)/(18))) =sin(π +(π/2)−((7π)/(18)))=−sin((π/2)−((7π)/(18)))=−cos(((7π)/(18))) ⇒ cos(70^o ) +cos(200^o )=0

$$\mathrm{70}^{{o}} =\frac{\mathrm{70}\pi}{\mathrm{180}}\:=\frac{\mathrm{7}\pi}{\mathrm{18}}\:\:\:{and}\:\mathrm{200}^{{o}} =\frac{\mathrm{200}\pi}{\mathrm{180}}\:=\frac{\mathrm{20}\pi}{\mathrm{18}}\:\Rightarrow \\ $$$${cos}\left(\mathrm{70}^{{o}} \right)+{sin}\left(\mathrm{200}^{{o}} \right)={cos}\left(\frac{\mathrm{7}\pi}{\mathrm{18}}\right)+{sin}\left(\frac{\mathrm{20}\pi}{\mathrm{18}}\right) \\ $$$$\frac{\mathrm{7}\pi}{\mathrm{18}}\:+\frac{\mathrm{20}\pi}{\mathrm{18}}\:=\frac{\mathrm{27}\pi}{\mathrm{18}}\:=\frac{\mathrm{3}\pi}{\mathrm{2}}\:\Rightarrow{sin}\left(\frac{\mathrm{20}\pi}{\mathrm{18}}\right)={sin}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}−\frac{\mathrm{7}\pi}{\mathrm{18}}\right) \\ $$$$={sin}\left(\pi\:+\frac{\pi}{\mathrm{2}}−\frac{\mathrm{7}\pi}{\mathrm{18}}\right)=−{sin}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{7}\pi}{\mathrm{18}}\right)=−{cos}\left(\frac{\mathrm{7}\pi}{\mathrm{18}}\right)\:\Rightarrow \\ $$$${cos}\left(\mathrm{70}^{{o}} \right)\:+{cos}\left(\mathrm{200}^{{o}} \right)=\mathrm{0} \\ $$

Answered by ajfour last updated on 15/Nov/19