My Math Forum Angle conversion

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 November 2nd, 2018, 11:05 PM #1 Member   Joined: Aug 2017 From: India Posts: 48 Thanks: 2 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.
 November 2nd, 2018, 11:53 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,172 Thanks: 1142 \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*}
November 3rd, 2018, 12:45 AM   #3
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Quote:
 Originally Posted by MathsLearner123 . . . when represented as 360 in the net . . .
What does "360 in the net" mean?

November 3rd, 2018, 10:05 AM   #4
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Quote:
 Originally Posted by skipjack What does "360 in the net" mean?
an online solution?

in the "internet" ?

 November 3rd, 2018, 11:25 AM #5 Global Moderator   Joined: Dec 2006 Posts: 19,878 Thanks: 1835 Perhaps, but I tried it on some websites and they got it right.
 November 3rd, 2018, 11:32 AM #6 Senior Member     Joined: Sep 2015 From: USA Posts: 2,172 Thanks: 1142 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 $-60^\circ + 180^\circ = 120^\circ$ Thanks from MathsLearner123
 November 3rd, 2018, 11:43 AM #7 Global Moderator   Joined: Dec 2006 Posts: 19,878 Thanks: 1835 It's easier to understand by considering the geometric definition of tan. Thanks from greg1313
November 3rd, 2018, 06:18 PM   #8
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Quote:
 Originally Posted by skipjack It's easier to understand by considering the geometric definition of tan.
Can you please explain me the method ?

 November 4th, 2018, 03:56 AM #9 Global Moderator   Joined: Dec 2006 Posts: 19,878 Thanks: 1835 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$^\circ$). Thanks from MathsLearner123

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