7:10 But θ=-1.176 is not in the third quadrant. The inverse sin we get for -12/13 is in the fourth quadrant but we need the one outside of our interval [-π/2:π/2]. When drawing the unit circle, it is clear that we are missing -π/2 and some angle to be in the third. This piece can be found by calculating sin(-12/13+π/2)=cos(12/13) and subtracting cos^-1(12/13) from -π/2 to get -1,96 [same as for the -cos^-1(-5/13)].
Zhu Ziyuan 0 seconds ago i reckon that by taking the arc tangent(12/5) to get theta+1.176, it would get you theta=1.176 like a reference angle. Using -(pi-reference angel) to get to the right angel in the targeted quadrant, the theta is -1.96 in Q3. From my point of view the answer is Z=3(COS(-1.96)+iSIN(-1.96))
My thought exactly. His answer aught to be 181.18 as far as I can tell. I've always had to reason my way through it based on the mod arg notation. I don't know of any way of working it out mathematically.
@@saneledegraaf6955 The issue seems to be that the inverse trigonometric functions are supposed to give out the minimal angle which results in such a value. But looking at the unit circle one sees that for each value of x or y there are two points and possible solutions, thus, the inverse functions cannot give us the complete answer and we have to rely on our reasoning of which angle satisfies both sin and cos functions.
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7:10 But θ=-1.176 is not in the third quadrant. The inverse sin we get for -12/13 is in the fourth quadrant but we need the one outside of our interval [-π/2:π/2].
When drawing the unit circle, it is clear that we are missing -π/2 and some angle to be in the third. This piece can be found by calculating sin(-12/13+π/2)=cos(12/13) and subtracting cos^-1(12/13) from -π/2 to get -1,96 [same as for the -cos^-1(-5/13)].
thank you so much!!! I was searching for this explanation, you commented really help me :)
I was going down to comment the same thing but you happened to do it 6 years before me.
1:02 "How do you see that in your head?" 🤣
Use of tan and arctangent are always appropriate and often simpler for students to understand then going back to sign divided by cosine.
I wish i had a teacher as passionate as you 🙌🏽 great work man
Zhu Ziyuan
0 seconds ago
i reckon that by taking the arc tangent(12/5) to get theta+1.176, it would get you theta=1.176 like a reference angle. Using -(pi-reference angel) to get to the right angel in the targeted quadrant, the theta is -1.96 in Q3. From my point of view the answer is Z=3(COS(-1.96)+iSIN(-1.96))
Just FYI, the problem at the very end. True answer should be for Arg(x)=-1.9656.
I used matlab and type in : angle(-5-12i)
Sir, I didn't understand how you calculated the angle Theta ?
This guys good!
Its in the forth quadrant, so one has to do 180-tan^-1(sin[θ]/cos[θ]) right?
No, then you get it in the first quadrant.
You should always diagram your complex number at the very beginning of your approach to the problem.
The answer at the end is in the fourth quadrant so is wrong
My thought exactly. His answer aught to be 181.18 as far as I can tell. I've always had to reason my way through it based on the mod arg notation. I don't know of any way of working it out mathematically.
@@saneledegraaf6955 The issue seems to be that the inverse trigonometric functions are supposed to give out the minimal angle which results in such a value. But looking at the unit circle one sees that for each value of x or y there are two points and possible solutions, thus, the inverse functions cannot give us the complete answer and we have to rely on our reasoning of which angle satisfies both sin and cos functions.
The last angle should have been (-pi) + (arctan 12/5) = 1.902