三角比の基本定理
\[ 0\tcdegree \leqq \theta \leqq 90\tcdegree \] $ \sin(90\tcdegree - \theta)$ = $\cos \theta $ ○ × Click! Anser$ \cos(90\tcdegree - \theta)$ = $\sin \theta $ ○ × Click! Anser
$ \tan(90\tcdegree - \theta)$ = $\displaystyle \frac{1}{\tan \theta} $ ○ × Click! Anser
\[ 0\tcdegree \leqq \theta \leqq 180\tcdegree \] $ \sin(180\tcdegree - \theta)$ = $\sin \theta $ ○ × Click! Anser
$ \cos(180\tcdegree - \theta)$ = $-\cos \theta $ ○ × Click! Anser
$ \tan(180\tcdegree - \theta)$ = $-\tan \theta ( \theta \neq 90\tcdegree ) $ ○ × Click! Anser
\[ 0\tcdegree \leqq \theta \leqq 90\tcdegree \] $ \sin(90\tcdegree + \theta)$ = $\cos \theta $ ○ × Click! Anser
$ \cos(90\tcdegree + \theta)$ = $-\sin \theta $ ○ × Click! Anser
$ \tan(90\tcdegree + \theta)$ = $- \displaystyle \frac{1}{\tan \theta} ( \theta \neq 0\tcdegree , \theta \neq 90\tcdegree ) $ ○ × Click! Anser
\[ 0\tcdegree \leqq \theta \leqq 180\tcdegree \] $ \tan \theta = $ $ \displaystyle \frac{\sin \theta}{\cos \theta} ( \theta \neq 90\tcdegree ) $ ○ × Click! Anser
$ \sin^2 \theta + \cos^2 \theta$ = $ 1 $ ○ × Click! Anser
$ 1 + \tan^2 \theta$ = $\displaystyle \frac{1}{cos^2 \theta} ( \theta \neq 90\tcdegree ) $ ○ × Click! Anser
直線 $ y = mx $ の x軸との角を $ \theta ( 0\tcdegree \leqq \theta \leqq 180\tcdegree , \theta \neq 90\tcdegree ) $ とすると、
$ m $ = $ \tan \theta $ ○ × Click! Anser
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平均点:10.00
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三角関数の加法定理 と その語呂合わせ |