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[Group5][Complex]Exam Question Practice


tutorinseoul

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5 minutes ago, SC2Player said:

Answer in pdf form:

Answer.pdf

aha, very well done except for some mistake with algebra! Try again for the quadratic substitution and derive a correct answer! (The value was wrong.)
Also, for the last part, there could be more than one solution in the second quadrant. Ask me if you get stuck for the second part :)!

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On 3/22/2017 at 4:03 PM, tutorinseoul said:

 

aha, very well done except for some mistake with algebra! Try again for the quadratic substitution and derive a correct answer! (The value was wrong.)
Also, for the last part, there could be more than one solution in the second quadrant. Ask me if you get stuck for the second part :)!

Bleh I can't figure out the second part.  Can you give me a hint?  Also was busy in Korea this week, so couldn't answer your questions alas.  Ima do them now (most seem pretty interesting).  

Edited by SC2Player
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22 minutes ago, SC2Player said:

Bleh I can't figure out the second part.  Can you give me a hint?  Also was busy in Korea this week, so couldn't answer your questions alas.  Ima do them now (most seem pretty interesting).  

Hint:
What's cos (9 pi / 10)?

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25 minutes ago, SC2Player said:

Bleh I can't figure out the second part.  Can you give me a hint?  Also was busy in Korea this week, so couldn't answer your questions alas.  Ima do them now (most seem pretty interesting).  

My triick will be to compare with the closest angle I know! For example, we have to get cos 7pi/10. The closest angle I know is 5pi/10 with exact angle. cos(pi/2) being zero, and since it's decreasing till pi, we know for my y=cos(7pi/10) value it must be bounded by 0 and -1. And then I can compare the value. I will write it up nearly tonight! Been busy in Seoul too lol.

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22 minutes ago, SC2Player said:

My attempt at finding cos(7pi/10) is in the attached file, but I'm not too sure if it's quite what the question was looking for - feels a bit inefficient tbh.  Also had a typo in my original file (had 5sqrt(5) instead of sqrt(5) for some reason), so that's probably where my value was wrong.  

Answer.pdf

Hey, I've written up my solution; see if you can understand and ask me if there unclear part!

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23 minutes ago, SC2Player said:

My attempt at finding cos(7pi/10) is in the attached file, but I'm not too sure if it's quite what the question was looking for - feels a bit inefficient tbh.  Also had a typo in my original file (had 5sqrt(5) instead of sqrt(5) for some reason), so that's probably where my value was wrong.  

Answer.pdf

Link

http://m.imgur.com/Rxpgevz,rhLNNT5,tO2yIVu

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@tutorinseoul I think the testing values is unnecessary. 
 

Spoiler

cos θ = ± sqrt((5 ± sqrt (5)) / 8) or 0. Note that there are 5 values of cos, and as expected, 5 angles from 0 to pi that satisfy cos (5θ) = 0, which are π/10, 3π/10, π/2, 7π/10, and 9π/10. Since the cos curve is decreasing from 0 to π, cos (7π/10) gets the second smallest (second least positive) value, which is -sqrt((5 - sqrt (5)) / 8). Therefore in finding values of cos π/10, it is also important to state that because cos π/10 >cos 3π/10, it is the largest of the 5 cos values. 

@SC2Player. You are right. The typo you made mistakenly allowed you to reject 5 - 5sqrt (5)) / 8 as a value of cos2θ because it is negative. It is always important to make sure the roots you reject actually can be rejected. 

Edited by kw0573
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1 hour ago, kw0573 said:

@tutorinseoul I think the testing values is unnecessary. 
 

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cos θ = ± sqrt((5 ± sqrt (5)) / 8) or 0. Note that there are 5 values of cos, and as expected, 5 angles from 0 to pi that satisfy cos (5θ) = 0, which are π/10, 3π/10, π/2, 7π/10, and 9π/10. Since the cos curve is decreasing from 0 to π, cos (7π/10) gets the second smallest (second least positive) value, which is -sqrt((5 - sqrt (5)) / 8). Therefore in finding values of cos π/10, it is also important to state that because cos π/10 >cos 3π/10, it is the largest of the 5 cos values. 

@SC2Player. You are right. The typo you made mistakenly allowed you to reject 5 - 5sqrt (5)) / 8 as a value of cos2θ because it is negative. It is always important to make sure the roots you reject actually can be rejected. 

You are right :)! I just ne d to think of pi, 3pi, 5 pi and 7pi over pi for my solutions as they only make the equation to be zero when they were substituted into cos 5x. Just wanted to use different approach :)! Thank you for your comments. 

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