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The book that i am reading shows an image similar to this: enter image description here

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 book

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  • $\begingroup$ The point is in the second quadrant, right? Is cosine positive or negative in the 2nd quadrant? Is $x$ positive or negative in the second quadrant? $\endgroup$ Commented Aug 7, 2015 at 12:57
  • $\begingroup$ Yes, x is in second quadrant. Oh.. Actually they specify the coordinate as x not -x. But then, how do you prove that identity holds for all theta, if you actually make x to be positive in second quadrant.. (Because they say in the end, therefore identity holds for all theta) $\endgroup$
    – nz_
    Commented Aug 7, 2015 at 14:20
  • $\begingroup$ @isquared-KeepitReal - But x is not positive in the second quadrant. If you want the coordinates in the second quadrant to be (-x, y), that's OK, but then all of the formulas need to be changed accordingly. It will work but it's a bookkeeping mess. $\endgroup$ Commented Aug 7, 2015 at 14:40
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    $\begingroup$ If $P$ is in the first quadrant, then $x,y,\sin\theta,\cos\theta,\tan\theta$ are all positive, and the identity holds. If $P$ is in the 2nd quadrant, then $y,\sin\theta$ are positive, $x,\cos\theta,\tan\theta$ are negative, and the identity holds. If $P$ is in the 3rd quadrant, then $\tan\theta$ is positive, $x,y,\sin\theta,\cos\theta$ are all negative, and the identity holds. If $P$ is in the 4th quadrant...details left to the reader. Bottom line: no matter where $P$ is, no matter what $\theta$ is (well, $\theta\ne\pi/2,3\pi/2$), the identity holds. $\endgroup$ Commented Aug 7, 2015 at 23:00
  • $\begingroup$ Do you understand it now, i squared? $\endgroup$ Commented Aug 9, 2015 at 0:41

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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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