Can somebody explain to me graphically or intuitively why this is true? Because if we think in terms of the base of the triangle formed inside a unit circle it seems like it should not be negative.
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"it seems like it should me negative": I don't follow you. Flipping the triangle upside-down doesn't make its base negative. Can you sketch that ? – Apr 27 '16 at 07:45
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Typo there, sorry. – Anonymous Apr 27 '16 at 07:47
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Better to ask what $-x$ might mean itself in my opinion. An angle of $-x$ can be viewed as a clockwise rotation whereas $x$ is anticlockwise. Now in a unit circle you can record the $\cos$ of an angle as the horizontal distance on the $x$ axis and so intuitively the right triangle formed by rotating clockwise will have the same horizontal distance as one rotated anticlockwise.
Hope this helps.
Karl
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See we can write it as $\cos(0-x)$ so now use $\cos(a-b)=cos(a)cos(b)+sin(a)sin(b)$ to get the desired result
Archis Welankar
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Look at this figure. For an angle of $-\theta$, you mirror the figure vertically: you see that this does not change $\cos(\theta)$, but it does change the sign of $\sin(\theta)$
Wouter
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Intuitively $\cos(-\theta)$ measures the $x$-coordinate of a vector that measures $\theta$ degrees below the positive $x$-axis, so this is why we have $\cos(-\theta) = \cos\theta$.
Another way of seeing this is through the series representation of $\cos x$ given by
$$ \cos(-x) \;\; =\;\; 1 - \frac{(-x)^2}{2!} + \frac{(-x)^4}{4!} - \frac{(-x)^6}{6!} + \ldots \;\; =\;\; \cos x. $$
Mnifldz
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