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Let us assume that we are given an unknown angle $x$. We dont know if it is negative or positive, we only know its in the 4th quadrant. Also, we're given a trigonometric function of $x$ (for example $\tan(x)$). Now let's say we have to find $\cos(x/2)$. The problem here is, if $x$ is a positive angle, less than $360^{\circ}$, then $x/2$ lies in the 2nd quadrant and is negative. If it's a negative angle less than $90^{\circ}$, then it lies in the 4th quadrant and is positive. How do we decide what sign to use? Also, for positive values greater than $360^{\circ}$, how do we find out what quadrant it lies in? (if $x=560^{\circ}$, $x/2$ lies in the 3rd quadrant and so forth). Is there any way to definitively decode or do we just write $\pm$ value of cos ?

Please note that i just graduated into 11th grade and have only a rudimentary understanding of trigonometry. I ask that everyone keep their answer simple and that they dont make it too technical.

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If $\displaystyle270^\circ<x<360^\circ,135^\circ<\frac x2<180^\circ\implies\cos\frac x2<0$

Also $\displaystyle\cos x>0\implies\cos x=+\sqrt{\frac1{1+\tan^2x}}$

and $\displaystyle\cos\frac x2=-\sqrt{\frac{1+\cos x}2}$ as $\displaystyle\cos2y=2\cos^2y-1$

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  • $\begingroup$ This is assuming that x is positive and lies between 270 and 360. Is x was say, 280, what you said would be true, but if x was 620, then the value of cos(x/2) would be wrong. I understood the identity part, that is to find the value, what im concerned about is the sign that goes with it. $\endgroup$ – t3hCrush3r Aug 13 '14 at 13:32
  • $\begingroup$ $x=620$ doesn't lie in the 4th quadrant. Use the formula $\cos(x-360^{\circ})=\cos(x)$ $\endgroup$ – rae306 Aug 13 '14 at 13:36
  • $\begingroup$ Typo, i meant- 680, or say 700. (700 = 360+340) lies in the 4th quadrant. But, then, x/2 (where x is 700) would be 350, and ALSO be in the 4th quadrant, making it positive. $\endgroup$ – t3hCrush3r Aug 13 '14 at 15:36
  • $\begingroup$ @t3hCrush3r, You may mean $$360^\circ+270^\circ<x<360^\circ+360^\circ\iff315^\circ<\frac x2<360^\circ\implies\cos\frac x2>0$$ $\endgroup$ – lab bhattacharjee Aug 13 '14 at 16:16
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Both $180$ and $360$ are on the boundary lines of quadrants.

We really do not need to assign a specific quadrant to $180$ or $360$ to get $\cos (180)=-1$ or $\cos (360)=1$

Note that $f(x) = \cos(x)$ is a continuous function and the value of the function at any $x$ is the same as the right or the left limit of the function at that point.

Thus no matter which quadrant you use to approach $180$ or $360$ you still get the same answer.

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