I want to know how can we imagine an angle which more than $360^\circ$. Since when it reaches its initial side after completing one rotation of $360^\circ$ degrees ($\sin 420^\circ = \sin60^\circ$).
Why do we need angles more than $360^\circ$, and how are they useful in real life?
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$\begingroup$Why do we need angles more than $360$ degrees and how are they useful in real life?
Looking at the problem through trigonometry glasses $(\mod 360^{\circ})$ of course you can't see needs using angles greater than $360^{\circ}$. But physically such angles make sense in concepts such as:
- Angular acceleration
- Angular velocity
- Angular displacement
- And many other physical concepts: angular position of the shaft, euler angles,...
It is also worth understanding that angles are just an idea of dividing a circle. It does not matter if we use degrees measures or other measures such as Arc measure of an angle or Gradian
$\endgroup$ $\begingroup$Consider a non-zero complex number $a + bi = (\sqrt{a^2 + b^2})[\frac a{\sqrt{a^2 + b^2}} + \frac b{\sqrt{a^2 + b^2}} i ] = (\sqrt{a^2 + b^2})[\cos \arctan \frac ba + i \sin \arctan \frac ba ]=r(\cos \theta + i \sin \theta)$ where $r = \sqrt{a^2 + b^2}$ and $\theta = \arctan \frac ba$.
Now consider another non-zero complex number $c + di = s (\cos \phi + i \sin \phi)$ where $s =\sqrt{c^2 + d^2}$ and $\phi = \arctan \frac ab$.
Then $(a+bi)(c+di) = (ac -bd) + (bc+ad)i$ is equal to $r(\cos \theta + i \sin \theta) s (\cos \phi + i \sin \phi)=rs (\cos (\theta + \phi) + i\sin(\theta + \phi)$
And in particular this means that $(a+ bi)^k = r^k(\cos \theta + i\sin \theta)^k=r^k(\cos (k\theta) + i\sin (k\theta))$.
For example take $z = -1 +\sqrt 3 i$. Now figure ought what $z^5$ is. We could to it the hard way.
$(-1 + \sqrt 3 i)^5 =-1^5+ 5*1^4 \sqrt 3 i - 10*1^3*(\sqrt 3 i)^2+ 10*1^2(\sqrt 3 i)^3- 5*1(\sqrt 3 i)^4 + (\sqrt 3 i)^5=$
$-1 + 5\sqrt 3i+ 10*3 -10*3*\sqrt 3i -5*9+9\sqrt 3 i=$
$-16 -16\sqrt 3i$.
Or we could figure $z = -1+ \sqrt 3i = 2(-\frac 12 + i\frac {\sqrt 3}2) = 2(\cos 120^\circ +i\sin 120^\circ)$ and $z^5 = 2^5(\cos (5\times 120) +i (\sin(5\times 120)) = 32(\cos 600 + i \sin 600)$.
And $600 = 360 + 240$ and $\cos 600 = \cos 240 = -\frac 12$ and $\sin 600 = \sin 240 = -\frac {\sqrt 3}2$ and so $z^5 = 32 (-\frac 12 - \frac {\sqrt 3}2i) =-16 - 16\sqrt {3} i$.
So that is one reason it is useful to consider angles over $360$.
Furthermore: Suppose I asked you to solve $w^5 = -16 - 16\sqrt {3}i$.
Wel you'd figure $w^5 = 32 (-\frac 12 -\frac {\sqrt 3}2 i) = 32(\cos 240 + i\sin 240)$
So we figure $w = \sqrt[5]{32}(\cos \frac {240}5 + i\sin \frac {240}5)=2(\cos 48 + i \sin 48)$
And indeed that is one answer. But as we saw $z = -1 + i \sqrt 3$ was another answer. And that makes sense $w^5 - (-16 - 16\sqrt {3}i)0$ ought to have FIVE answers. $2(\cos 48 + i\sin 48)$ is one of them, and $-1 + i \sqrt 3$ is another. But what are the remaining three?
Well, use angles larger than $360$.
$w^5 = 32 (-\frac 12 -\frac {\sqrt 3}2 i) = 32(\cos 240 + i\sin 240)=$
$32=(\cos 600 + i\sin 600)$ so $w_2 = \sqrt[5]{32}(\cos \frac{600}{5} + i\sin \frac {600}{5})= 2(\cos 120 + i \sin 120) = -1 +\sqrt 3 i$ is another (we already had this one).
And $w^5 = 32(\cos 960 + i\sin 960)$ so $w_3=2(\cos 192 + i \sin 192)$ is a third.
And $w^5= 32(\cos 1320 + i\sin 1320)$ so $w_4 = 2(\cos 264 + i\cos 264)$ is a fourth.
And finally $w^5 = 32(\cos 1680 + i \sin 1680)$ so $w_5 = 2(\cos 336 + i \sin 336)$ is the fifth root.
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