I want to know if,
$-\cos(t) = \cos(t+180)$
or
$-\cos(t) = \cos(t-180)$
Please guide me. Thanks
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See Shift by area – Amzoti Apr 13 '13 at 03:13
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1Both. A 180 degree shift in a sinusoidal function is indistinguishable from a phase reversal. – Kaz Apr 13 '13 at 05:42
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Use the degree symbol: "^\circ". If you don't use it, you are using radians. – Stefan Smith Apr 13 '13 at 17:20
3 Answers
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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:
amWhy
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2@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. – Mikasa Apr 13 '13 at 04:58
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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 $\sin(t+180^\circ)=-\sin t$.
André Nicolas
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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.
colormegone
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