# What is -cos(t) equivalent to in terms of cos(t)

I want to know if,

$-\cos(t) = \cos(t+180)$

or

$-\cos(t) = \cos(t-180)$

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– Amzoti Apr 13 '13 at 3:13
Both. A 180 degree shift in a sinusoidal function is indistinguishable from a phase reversal. – Kaz Apr 13 '13 at 5:42
Use the degree symbol: "^\circ". If you don't use it, you are using radians. – Stefan Smith Apr 13 '13 at 17:20

Yes, both are true: $$-\cos (t) = \cos (t \pm 180^\circ)\quad\text{ or in radians, }\;\;-\cos(t) = (t \pm \pi)$$

Note that $$t + 180^\circ - (t - 180^\circ) = 360^\circ$$ See this link for similar trigonometric "shifts"

In the same Wikipedia article, you'll find a handy diagram with ordered pairs $(\cos x, \sin x)$ for angle x measured in radians, as they appear cycle the unit circle:

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You guys type fast! +1 – Amzoti Apr 13 '13 at 3:26
@Amzoti: Yes, they are toooo fast. and this is why I cannot make my approach as soon as possible. Nice to see you here. +1 for being fast. – Babak S. Apr 13 '13 at 4:58
@BabakS.: Hello my friend - good to see you again! – Amzoti Apr 13 '13 at 5:01

Both answers are correct. Note that $t+180^\circ$ and $t-180^\circ$ differ by $360^\circ$, so they have the same cosine (and sine).

Let's verify that $\cos(t+180^\circ)=-\cos t$. If you rotate a point $(x,y)$ around the origin through $180^\circ$, both the $x$ and $y$-coordinate of the point change sign. It follows that $\cos(t+180^\circ)=-\cos t$ and $\sin(t+180^\circ)=-\sin t$.

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As a separate verification, you can use the "angle-addition" formula for cosine:

$$\cos (t \pm 180^{\circ}) = \cos t \cdot \cos (\pm 180^{\circ}) \mp \sin t \cdot \sin (\pm 180^{\circ})$$

$$= (\cos t) \cdot [-1] \mp (\sin t) \cdot 0 = -\cos t .$$

You can reconstruct a lot of trig identities for complementary angles, supplementary angles, etc. with the angle-addition rules for sine, cosine, and tangent.

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