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November 2nd, 2018, 11:05 PM   #1
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Angle conversion

I calculated arc tan or inverse tan of sqrt(-3) and the result is -60. But when represented as 360 in the net it is shown as 120 Degrees. How -60 Deg is same as 120 Deg? Please advise. I know it will be something like 180 - 60 = 120. But why I should take 180 degrees and subtract 60? How that symmetry is coming?

Last edited by skipjack; November 3rd, 2018 at 12:45 AM.
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November 2nd, 2018, 11:53 PM   #2
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$\begin{align*}

&\tan(120^\circ) = \\

&\tan(-60^\circ + 180^\circ) = \\

&\dfrac{\sin(-60^\circ+180^\circ)}{\cos(-60^\circ + 180^\circ)} = \\

&\dfrac{\sin(-60^\circ)\cos(180^\circ)+\cos(-60^\circ)\sin(180^\circ)}
{\cos(-60^\circ)\cos(180^\circ)-\sin(-60^\circ)\sin(180^\circ)} = \\

&\dfrac{-\sin(-60^\circ)}{-\cos(-60^\circ)} =\\

&\dfrac{\sin(-60^\circ)}{\cos(-60^\circ)} = \\

&\tan(-60^\circ)

\end{align*}$
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November 3rd, 2018, 12:45 AM   #3
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Quote:
Originally Posted by MathsLearner123 View Post
. . . when represented as 360 in the net . . .
What does "360 in the net" mean?
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November 3rd, 2018, 10:05 AM   #4
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Quote:
Originally Posted by skipjack View Post
What does "360 in the net" mean?
an online solution?

in the "internet" ?
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November 3rd, 2018, 11:25 AM   #5
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Perhaps, but I tried it on some websites and they got it right.
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November 3rd, 2018, 11:32 AM   #6
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Basically the idea is that that the tangent is the ratio of the sine and cosine functions and will remain the same if the signs of both of those are flipped.

$\sin(\theta) = -\sin(\theta + 180^\circ)$

$\cos(\theta) = -\cos(\theta + 180^\circ)$

and thus

$\tan(\theta) = \tan(\theta + 180^\circ)$

In order to convert from a domain of say $(-180^\circ, 180^\circ]$ to $[0, 360)$

we simply add $180^\circ$ to the angle.

In this case $-60^\circ + 180^\circ = 120^\circ$
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November 3rd, 2018, 11:43 AM   #7
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It's easier to understand by considering the geometric definition of tan.
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November 3rd, 2018, 06:18 PM   #8
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Quote:
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It's easier to understand by considering the geometric definition of tan.
Can you please explain me the method ?
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November 4th, 2018, 03:56 AM   #9
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ThetaCircle.PNG
If, in the above diagram, x is non-zero, tan(θ) = y/x. If the point (x, y) is moved to a different position on the circumference of the circle, so that its coordinates become (-x, -y), the ratio of those coordinates is unchanged. The point's movement can be thought of as due to a rotation of the triangle through ±180$^\circ$, so that the angle θ becomes θ ± 180$^\circ$.

The tangent of the angle is unchanged, i.e. y/x = tan(θ) = tan(θ ± 180$^\circ$).
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