Question Number 99321 by Ar Brandon last updated on 20/Jun/20
![[((Angle(𝛉)),(sin(𝛉)),(cos(𝛉)[)),(0^° ,0,1),((15°),(((√6)−(√2))/4),(((√6)+(√2))/4)),((18°),(((√5)−1)/4),((√(5+(√5)))/(2(√2)))),((30°),(1/2),((√3)/2)),((36°),((√(5−(√5)))/(2(√2))),(((√5)+1)/4)),((45°),(1/(√2)),(1/(√2))),((54°),(((√5)+1)/4),((√(5−(√5)))/(2(√2)))),((60°),((√3)/2),(1/2)),((72°),((√(5+(√5)))/(2(√2))),(((√5)−1)/4)),((75°),(((√6)+(√2))/4),(((√6)−(√2))/4)),((90°),1,0) ] sin(𝛉)=cos(90°−𝛉)](Q99321.png)
$$\begin{bmatrix}{\boldsymbol{\mathrm{Angle}}\left(\boldsymbol{\theta}\right)}&{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\theta}\right)}&{\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\theta}\right)\left[\right.}\\{\mathrm{0}^{°} }&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{15}°}&{\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}}}&{\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}}\\{\mathrm{18}°}&{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}}&{\frac{\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}}{\mathrm{2}\sqrt{\mathrm{2}}}}\\{\mathrm{30}°}&{\frac{\mathrm{1}}{\mathrm{2}}}&{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\\{\mathrm{36}°}&{\frac{\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}{\mathrm{2}\sqrt{\mathrm{2}}}}&{\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}}\\{\mathrm{45}°}&{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}}&{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}}\\{\mathrm{54}°}&{\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}}&{\frac{\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}{\mathrm{2}\sqrt{\mathrm{2}}}}\\{\mathrm{60}°}&{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}&{\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{72}°}&{\frac{\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}}{\mathrm{2}\sqrt{\mathrm{2}}}}&{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}}\\{\mathrm{75}°}&{\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}}&{\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}}}\\{\mathrm{90}°}&{\mathrm{1}}&{\mathrm{0}}\end{bmatrix} \\ $$$$\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\theta}\right)=\boldsymbol{\mathrm{cos}}\left(\mathrm{90}°−\boldsymbol{\theta}\right) \\ $$
Commented by bobhans last updated on 20/Jun/20
$$\mathrm{waw}...\mathrm{cooll} \\ $$
Commented by mahdi last updated on 20/Jun/20
$$\left.:\right) \\ $$
Answered by MWSuSon last updated on 20/Jun/20
$$\mathrm{This}\:\mathrm{is}\:\mathrm{really}\:\mathrm{helpful}\:\mathrm{sir}. \\ $$
Commented by bobhans last updated on 20/Jun/20
$$\mathrm{yes}\:\mathrm{sir} \\ $$
Commented by Ar Brandon last updated on 20/Jun/20
Thanks��