0度と135度の三角比(sin, cos, tan)の値を求める問題です。表の空欄を埋めます。幾何学三角比三角関数sincostan角度2025/7/111. 問題の内容0度と135度の三角比(sin, cos, tan)の値を求める問題です。表の空欄を埋めます。2. 解き方の手順* 0度の三角比 * sin(0∘)=0sin(0^\circ) = 0sin(0∘)=0 * cos(0∘)=1cos(0^\circ) = 1cos(0∘)=1 * tan(0∘)=sin(0∘)cos(0∘)=01=0tan(0^\circ) = \frac{sin(0^\circ)}{cos(0^\circ)} = \frac{0}{1} = 0tan(0∘)=cos(0∘)sin(0∘)=10=0* 135度の三角比 * sin(135∘)=12=22sin(135^\circ) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}sin(135∘)=21=22 * cos(135∘)=−12=−22cos(135^\circ) = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}cos(135∘)=−21=−22 * tan(135∘)=sin(135∘)cos(135∘)=12−12=−1tan(135^\circ) = \frac{sin(135^\circ)}{cos(135^\circ)} = \frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}} = -1tan(135∘)=cos(135∘)sin(135∘)=−2121=−13. 最終的な答え* ア: 0* イ: 1* ウ: 0* エ: 2\sqrt{2}2* オ: 2\sqrt{2}2* カ: -1