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March 28, 2024, 10:53:39 pm

Author Topic: 3U Maths Question Thread  (Read 1230227 times)  Share 

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anotherworld2b

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Re: 3U Maths Question Thread
« Reply #1230 on: January 21, 2017, 10:43:45 pm »
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I'm confused on how to use the diagram to find what is necessary to find the exact values for sin (a+b) , cos (A-B) and tan (A+B)
What happened to negative for Tan B = -3/4 in the diagram?
I had a another quick question to ask if you are required to find the exact value of 135° Would the answer be -cos (45) = 1/square root of 2?
I only just realised i marked my answers correct without knowing what the actual answers were so i wanted check if inwas right or not

 
These diagrams may be a tad confusing depending on how you have been taught to use the angles of any magnitude system, let me know! :)

(Image removed from quote.)
« Last Edit: January 21, 2017, 10:48:01 pm by anotherworld2b »

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #1231 on: January 22, 2017, 12:44:17 am »
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I'm confused on how to use the diagram to find what is necessary to find the exact values for sin (a+b) , cos (A-B) and tan (A+B)
What happened to negative for Tan B = -3/4 in the diagram?
I had a another quick question to ask if you are required to find the exact value of 135° Would the answer be -cos (45) = 1/square root of 2?
I only just realised i marked my answers correct without knowing what the actual answers were so i wanted check if inwas right or not

The basic idea is to use Pythagoras theorem to find the missing side of the triangles, and then use angle expansion formulae to figure out the answer. Do you learn, for example:



You can find the values of \(\sin{a},\cos{a},\sin{b}\) (etc) by using right angled trigonometry with those diagrams (once you've found the missing side using Pythagoras theorem).

Yep, you're totally right, the bottom side should be -4 in the diagram, not 4, sorry :)

If you mean the exact value of \(\cos{135}\), then yep!


Rathin

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Re: 3U Maths Question Thread
« Reply #1232 on: January 22, 2017, 12:31:35 pm »
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Given dr/dθ = 3/(1-r)^4 and r(0)=0, use the substitution u=1-r to find r in terms of θ.
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #1233 on: January 22, 2017, 02:15:17 pm »
+1
Given dr/dθ = 3/(1-r)^4 and r(0)=0, use the substitution u=1-r to find r in terms of θ.

Hmmm, that's a really tough one. I've worked through a solution for you, every step should be reasonably clear, but definitely get this clarified if you need it, or if you're not sure why I did something. It involves a fair bit of manipulating your derivatives. VERY tough, would be one of the last questions in a HSC exam ;D

SOLUTION


Rathin

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Re: 3U Maths Question Thread
« Reply #1234 on: January 22, 2017, 03:18:58 pm »
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Hmmm, that's a really tough one. I've worked through a solution for you, every step should be reasonably clear, but definitely get this clarified if you need it, or if you're not sure why I did something. It involves a fair bit of manipulating your derivatives. VERY tough, would be one of the last questions in a HSC exam ;D

SOLUTION


Genius! Thanks :)
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hanaacdr

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Re: 3U Maths Question Thread
« Reply #1235 on: January 22, 2017, 04:15:56 pm »
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hi
Can i please get some help on this question
sin4x= 2sin2x

and would it be easier to change is to cos instead
thanks

hanaacdr

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Re: 3U Maths Question Thread
« Reply #1236 on: January 22, 2017, 04:18:56 pm »
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also sorry
but how do you do this question,
sinx = tanx

thank you!

Rathin

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Re: 3U Maths Question Thread
« Reply #1237 on: January 22, 2017, 04:43:38 pm »
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hi
Can i please get some help on this question
sin4x= 2sin2x

and would it be easier to change is to cos instead
thanks

sin(4x)=2sin(2x)
∴sin(2x)cos(2x)=2sin(2x)
∴sin(2x)cos(2x)-2sin(2x)=0
∴cos(2x)-sin(2x)=0
∴sqrt(2)cos(2x-π/4)=0
∴2x-π/4=π/2
∴2x-π/4=2nπ+-π/2 where n belongs to all integers
∴2x=2nπ+-3π/4
∴x=nπ+-3π/8
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Rathin

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Re: 3U Maths Question Thread
« Reply #1238 on: January 22, 2017, 04:48:04 pm »
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also sorry
but how do you do this question,
sinx = tanx

thank you!

sin(x)=tan(x)
∴sin(x)=sin(x)/cos(x)
∴sin(x)cos(x)=sin(x)
∴sin(x)cos(x)-sin(x)=0
∴sin(x)(cos(x)-1)=0
∴sin(x)=0 or cos(x)=1
∴x=nπ or x= 2nπ ,where n belongs to all integers
« Last Edit: January 22, 2017, 04:51:19 pm by Rathin »
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hanaacdr

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Re: 3U Maths Question Thread
« Reply #1239 on: January 22, 2017, 04:50:18 pm »
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thank youuu
much appreciated!

sin(4x)=2sin(2x)
∴sin(2x)cos(2x)=2sin(2x)
∴sin(2x)cos(2x)-2sin(2x)=0
∴cos(2x)-sin(2x)=0
∴sqrt(2)cos(2x-π/4)=0
∴2x-π/4=π/2
∴2x-π/4=2nπ+-π/2 where n belongs to all integers
∴2x=2nπ+-3π/4
∴x=nπ+-3π/8

anotherworld2b

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Re: 3U Maths Question Thread
« Reply #1240 on: January 22, 2017, 05:36:43 pm »
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how do you know which side is negative for tan B -3/4?

The basic idea is to use Pythagoras theorem to find the missing side of the triangles, and then use angle expansion formulae to figure out the answer. Do you learn, for example:



You can find the values of \(\sin{a},\cos{a},\sin{b}\) (etc) by using right angled trigonometry with those diagrams (once you've found the missing side using Pythagoras theorem).

Yep, you're totally right, the bottom side should be -4 in the diagram, not 4, sorry :)

If you mean the exact value of \(\cos{135}\), then yep!



kiwiberry

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Re: 3U Maths Question Thread
« Reply #1241 on: January 22, 2017, 06:23:01 pm »
+1
how do you know which side is negative for tan B -3/4?

B must be either in the 2nd or 4th quadrants as these are where tan is negative. Since the question states that B is an obtuse angle, it must be in the 2nd quadrant. Therefore, the x-coordinate will be negative (-4) and the y-coordinate will be positive (3), making the 4 on the bottom of the triangle negative :)
« Last Edit: January 22, 2017, 06:48:42 pm by kiwiberry »
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #1242 on: January 22, 2017, 06:43:01 pm »
+1
Thanks to all the legendary helpers above, you guys are awesome ;D

anotherworld2b

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Re: 3U Maths Question Thread
« Reply #1243 on: January 22, 2017, 07:15:40 pm »
+1
That makes so much more sense :D
Thank you very much for you help :)

B must be either in the 2nd or 4th quadrants as these are where tan is negative. Since the question states that B is an obtuse angle, it must be in the 2nd quadrant. Therefore, the x-coordinate will be negative (-4) and the y-coordinate will be positive (3), making the 4 on the bottom of the triangle negative :)

bsdfjnlkasn

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Re: 3U Maths Question Thread
« Reply #1244 on: January 22, 2017, 08:46:35 pm »
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Hmmm, that's a really tough one. I've worked through a solution for you, every step should be reasonably clear, but definitely get this clarified if you need it, or if you're not sure why I did something. It involves a fair bit of manipulating your derivatives. VERY tough, would be one of the last questions in a HSC exam ;D

SOLUTION


Hey Jamon!

I don't understand how you got the first line arrow, it could be because i'm a bit rusty on my substitution method but would love an explanation regardless!

Thank you  :)

Edit: I also don't understand why the derivative expression needed to be flipped (next to the third therefore sign)  ???
« Last Edit: January 22, 2017, 08:50:56 pm by bsdfjn;lkasn »