The book that i am reading shows an image similar to this:
It then proceeds to state that $\sin \theta = \frac{y}{OP}$ and $\cos \theta = \frac{x}{OP}$ (where P is the point on that line with coordinates (x,y)). Therefore tangent is y/x for any angle theta. But, $\cos \theta = -\frac{x}{OP}$ and therefore $\tan \theta = -\frac{y}{x}$, no?
$$\sin(\pi - \alpha) \equiv \sin \theta \equiv \sin \alpha \equiv y/OP$$, but $$\cos(\pi - \alpha) \equiv \cos \theta \equiv -\cos \alpha \equiv -x/OP$$
Edit: Here is the photo
1 Answer
$\begingroup$Given the angle $\theta$, there are 4 possible scenarios (ignoring the cases when $\theta$ is a multiple of $\frac{\pi}{2}$ and $\pi$): \begin{eqnarray} \frac{\pi}{2} < \theta < 0 \\ \pi < \theta < \frac{\pi}{2} \\ \frac{3\pi}{2}< \theta < \pi \\ 2\pi < \theta < \frac{3\pi}{2} \\ \end{eqnarray}
These represent the quadrant in which $\theta$ resides. Quadrants I,II,III and IV respectively.
Also, imagine a point $P$ which lines on the line that creates an angle $\theta$ moving in the anti-clockwise manner. The general coordinates of point $P$ are $(x,y)$. Let $OP$ be the length of the aforementioned line. Which is always positive.
1) Given that $\theta$ is in I quadrant:
The coordinates of $P$ are $(x,y)$, hence $\sin(\theta) = \frac{y}{OP}$ and $\cos(\theta) = \frac{x}{OP}$, i.e. $\tan(\theta) = \frac{y}{x}$, $\tan(\theta) \equiv \frac{\sin(\theta)}{\cos(\theta)}$. Identity holds.
2) Given that $\theta$ is in II quadrant:
The coordinates of $P$ are $(-x,y)$, hence $\sin(\theta) = \frac{y}{OP}$ and $\cos(\theta) = \frac{-x}{OP}$, i.e. $\tan(\theta) = \frac{y}{-x}$, $-\tan(\theta) \equiv -\frac{\sin(\theta)}{\cos(\theta)}$. Identity holds. Or simply multiplying by $-1$ gives $\tan(\theta) \equiv \frac{\sin(\theta)}{\cos(\theta)}$.
3) Given that $\theta$ is in III quadrant:
The coordinates of $P$ are $(-x,-y)$, hence $\sin(\theta) = \frac{-y}{OP}$ and $\cos(\theta) = \frac{-x}{OP}$, i.e. $\tan(\theta) = \frac{-y}{-x}$, $\tan(\theta) \equiv \frac{\sin(\theta)}{\cos(\theta)}$. Identity holds.
4) Given that $\theta$ is in IV quadrant:
The coordinates of $P$ are $(x,-y)$, hence $\sin(\theta) = \frac{-y}{OP}$ and $\cos(\theta) = \frac{x}{OP}$, i.e. $\tan(\theta) = \frac{-y}{x}$, $-\tan(\theta) \equiv -\frac{\sin(\theta)}{\cos(\theta)}$. Identity holds. Or simply multiplying by $-1$ gives $\tan(\theta) \equiv \frac{\sin(\theta)}{\cos(\theta)}$.
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