00:00Okay, guys, root 2, cis, pi upon 6.
00:05Give it to solve and solve it.
00:07What does cis mean?
00:09If cis is pi by 6, then it's cos pi by 6 plus iota sin pi by 6.
00:14If cis is pi by 3, then cos is pi by 3, iota is sin pi by 3.
00:18So, it's pi by 6.
00:20Now, cos pi by 6, we know that root 3 by 2.
00:23Sin pi by 6, we know that half.
00:25This root 2 is either multiply or either.
00:29We know that root 2 into root 3, root 6.
00:31Here, root 2 and upon me, 2.
00:33LCM, how much?
00:35Root 6 plus iota root 2.
00:37This is the answer to this question.
00:40Here, cos 7 theta plus iota sin 7 theta, divide by cos 3 theta plus iota sin 3 theta.
00:49So, how can we do this?
00:51Now, I've told a demoverous theorem.
00:53I've told you.
00:55I've written that cos theta plus iota sin theta's power n is power n is power n.
01:00So, we can go down.
01:01Then, we can do it also.
01:02We can do it.
01:02So, cos n theta plus iota sin n theta is power n.
01:05We can write cos theta plus iota sin theta's power n.
01:07We can write it.
01:09This power angle 7 was power.
01:12I said, where will the power go from?
01:17Or where will the angle go from?
01:20Now use a little bit.
01:22If the basis is the same,
01:23the power goes down and subtracts
01:277-3-4
01:28Now let's multiply this 4
01:31and this is the answer to this question.
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