The point reached by an anticlockwise angle \(\theta\) on the unit circle is \((\cos\theta,\sin\theta)\). Clockwise angles are negative. Hence \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\), where \(\cos\theta\ne0\).
| Quadrant | Positive ratios |
|---|---|
| I | Sine, cosine and tangent |
| II | Sine |
| III | Tangent |
| IV | Cosine |
\(\sin(\pi-\theta)=\sin\theta,\quad\cos(\pi-\theta)=-\cos\theta,\quad\tan(\pi-\theta)=-\tan\theta\)
\(\sin(-\theta)=-\sin\theta,\quad\cos(-\theta)=\cos\theta,\quad\tan(-\theta)=-\tan\theta\)
A nonvertical line through the origin making angle \(\theta\) with the positive \(x\)-axis has gradient \(\tan\theta\) and equation \(y=x\tan\theta\). The angle is not itself the gradient.
For a point \(P(x,y)\ne O\) on the terminal ray, let \(r=\sqrt{x^2+y^2}>0\). Then \(\sin\theta=y/r\), \(\cos\theta=x/r\) and \(\tan\theta=y/x\) when \(x\ne0\). Keep the coordinate signs and use a positive radius.
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