• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Further Trig maths help (1 Viewer)

blackglitter

blackglitter

Member
Joined
May 25, 2008
Messages
198
Location
Somewhere on Earth
Gender
Female
HSC
2009
Hey guys, can you please help me with the following questions. Thanks

1) Show that
(2sin3A + 2 cos3 A) / (sin A + cos A ) = 2 – sin2A
if sin A + cos A does not equal 0


2) [sin A / (cos A + sin A) ] + [sin A / (cos A - sin A) = tan2A

3) (2tanA) / (1 + tan2A) = sin 2A

4) (sin 3A/sinA) - (cos3A/cosA) = 2
When sin A and cos A do not equal 0

Thank you :)
 
lyounamu

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
blackglitter said:
Hey guys, can you please help me with the following questions. Thanks

1) Show that
(2sin3A + 2 cos3 A) / (sin A + cos A ) = 2 – sin2A
if sin A + cos A does not equal 0


2) [sin A / (cos A + sin A) ] + [sin A / (cos A - sin A) = tan2A

3) (2tanA) / (1 + tan2A) = sin 2A

4) (sin 3A/sinA) - (cos3A/cosA) = 2
When sin A and cos A do not equal 0

Thank you :)
Click to expand...
Don't really refer to my answer for q1. It's just too absurd and ambiguous to great extent.

2(sin3a+cos3a) = 2(sin3a + sin(90-3a))= 2(sin45cos3a-45) ...(1)
sina + cosa = sina + sin(90-a) = sin45 cosa-45 ...(2)

(1)/(2):
2cos3a-45/cosa-45
= 2(cosa-45cos2a - sina-45sin2a)/cosa-45
= 2cos2a - tana-45sin2a
= 2cos2a - (tana - 1)/1+tana sin2a
= 2cos2a - (tana-1)/1+tana 2sinacosa
= 2cos2a - (2sin^2a - 2sinacosa)/1+tana
= 2cos2a - (2sin^2acosa - 2sinacos^2(a)/cosa+sina
= 2cos2a - (2(1-cos^2(a)cosa - 2sina (1-sin^2(a)/cosa+sina
= 2cos2a - (2cosa - 2cos^3(a) - 2sina + 2sin^3(a)/cosa+sina
= 2cos2a - (2(cosa-sina) - 2(cos^3(a) - sin^3(a))/cosa + sina
= 2cos2a - (2(cosa-sina)-2(cosa-sin)(cos^2(a) + cosasina + sin^2(a))/cosa + sina
= 2cos2a - (cosa-sina)(2-2(cos^2(a)+cosasina + sin^2(a))/cosa+sina
= 2cos2a - (cosa-sina)(2-2(1+cosasina))/cosa+sina
= 2cos2a - (cosa-sina) ( 2-2 - sin2a)/cosa+sina
= 2cos2a - (cosa-sina)(-sin2a)/cosa+sina
= 2cos2a - (cosa-sina)(cosa-sina) (-sin2a) / (cosa+sina)(cosa-sina)
= 2cos2a - (cos^2(a)-2sinacosa + sin^2(a))(-sin2a)/(cos^2(a)-sin^2(a)
= 2cos2a - (1-sin2a)(-sin2a)/cos2a
= 2cos2a - (-sin2a + sin^2(2a)/cos2a
= 2cos2a - (-tan2a + tan2asin2a)
= 2cos2a + tan2a + tan2asin2a
= 2cos^2(2a) + sin2a + sin^2(2a)
= 2(1-sin^2(2a)+sin2a+ sin^2(2a)
= 2 + sin2a - sin^2(2a)
= 2 + 2sinacosa - sin^2(2a)


My working-out is absolutely absurd. I know....After this, I don't feel like doing any maths...

EDIT: MY working out here will not get you anywhere. But I just realised that question is WRONG.
 
Last edited:
ratcher0071

ratcher0071

Member
Joined
Feb 17, 2008
Messages
617
Location
In Space
Gender
Male
HSC
2009
lyounamu said:
Don't really refer to my answer for q1. It's just too absurd and ambiguous to great extent.

2(sin3a+cos3a) = 2(sin3a + sin(90-3a))= 2(sin45cos3a-45) ...(1)
sina + cosa = sina + sin(90-a) = sin45 cosa-45 ...(2)

(1)/(2):
2cos3a-45/cosa-45
= 2(cosa-45cos2a - sina-45sin2a)/cosa-45
= 2cos2a - tana-45sin2a
= 2cos2a - (tana - 1)/1+tana sin2a
= 2cos2a - (tana-1)/1+tana 2sinacosa
= 2cos2a - (2sin^2a - 2sinacosa)/1+tana
= 2cos2a - (2sin^2acosa - 2sinacos^2(a)/cosa+sina
= 2cos2a - (2(1-cos^2(a)cosa - 2sina (1-sin^2(a)/cosa+sina
= 2cos2a - (2cosa - 2cos^3(a) - 2sina + 2sin^3(a)/cosa+sina
= 2cos2a - (2(cosa-sina) - 2(cos^3(a) - sin^3(a))/cosa + sina
= 2cos2a - (2(cosa-sina)-2(cosa-sin)(cos^2(a) + cosasina + sin^2(a))/cosa + sina
= 2cos2a - (cosa-sina)(2-2(cos^2(a)+cosasina + sin^2(a))/cosa+sina
= 2cos2a - (cosa-sina)(2-2(1+cosasina))/cosa+sina
= 2cos2a - (cosa-sina) ( 2-2 - sin2a)/cosa+sina
= 2cos2a - (cosa-sina)(-sin2a)/cosa+sina
= 2cos2a - (cosa-sina)(cosa-sina) (-sin2a) / (cosa+sina)(cosa-sina)
= 2cos2a - (cos^2(a)-2sinacosa + sin^2(a))(-sin2a)/(cos^2(a)-sin^2(a)
= 2cos2a - (1-sin2a)(-sin2a)/cos2a
= 2cos2a - (-sin2a + sin^2(2a)/cos2a
= 2cos2a - (-tan2a + tan2asin2a)
= 2cos2a + tan2a + tan2asin2a
= 2cos^2(2a) + sin2a + sin^2(2a)
= 2(1-sin^2(2a)+sin2a+ sin^2(2a)
= 2 + sin2a - sin^2(2a)
= 2 + 2sinacosa - sin^2(2a)
= 2 + 2sinacosa - 2(2sinacosa)
= 2 -2sinacosa
= 2-sin2a

My working-out is absolutely absurd. I know....After this, I don't feel like doing any maths...
Click to expand...
OMG !!!!
 
F

Forbidden.

Banned
Joined
Feb 28, 2006
Messages
4,436
Location
Deep trenches of burning HELL
Gender
Male
HSC
2007
lyounamu said:
Don't really refer to my answer for q1. It's just too absurd and ambiguous to great extent.

2(sin3a+cos3a) = 2(sin3a + sin(90-3a))= 2(sin45cos3a-45) ...(1)
sina + cosa = sina + sin(90-a) = sin45 cosa-45 ...(2)

(1)/(2):
2cos3a-45/cosa-45
= 2(cosa-45cos2a - sina-45sin2a)/cosa-45
= 2cos2a - tana-45sin2a
= 2cos2a - (tana - 1)/1+tana sin2a
= 2cos2a - (tana-1)/1+tana 2sinacosa
= 2cos2a - (2sin^2a - 2sinacosa)/1+tana
= 2cos2a - (2sin^2acosa - 2sinacos^2(a)/cosa+sina
= 2cos2a - (2(1-cos^2(a)cosa - 2sina (1-sin^2(a)/cosa+sina
= 2cos2a - (2cosa - 2cos^3(a) - 2sina + 2sin^3(a)/cosa+sina
= 2cos2a - (2(cosa-sina) - 2(cos^3(a) - sin^3(a))/cosa + sina
= 2cos2a - (2(cosa-sina)-2(cosa-sin)(cos^2(a) + cosasina + sin^2(a))/cosa + sina
= 2cos2a - (cosa-sina)(2-2(cos^2(a)+cosasina + sin^2(a))/cosa+sina
= 2cos2a - (cosa-sina)(2-2(1+cosasina))/cosa+sina
= 2cos2a - (cosa-sina) ( 2-2 - sin2a)/cosa+sina
= 2cos2a - (cosa-sina)(-sin2a)/cosa+sina
= 2cos2a - (cosa-sina)(cosa-sina) (-sin2a) / (cosa+sina)(cosa-sina)
= 2cos2a - (cos^2(a)-2sinacosa + sin^2(a))(-sin2a)/(cos^2(a)-sin^2(a)
= 2cos2a - (1-sin2a)(-sin2a)/cos2a
= 2cos2a - (-sin2a + sin^2(2a)/cos2a
= 2cos2a - (-tan2a + tan2asin2a)
= 2cos2a + tan2a + tan2asin2a
= 2cos^2(2a) + sin2a + sin^2(2a)
= 2(1-sin^2(2a)+sin2a+ sin^2(2a)
= 2 + sin2a - sin^2(2a)
= 2 + 2sinacosa - sin^2(2a)
= 2 + 2sinacosa - 2(2sinacosa)
= 2 -2sinacosa
= 2-sin2a

My working-out is absolutely absurd. I know....After this, I don't feel like doing any maths...
Click to expand...
You are absolutely wonderful I've always been the one notorious for using the most paper justified by the need to write step by step.

Rock On
 
ratcher0071

ratcher0071

Member
Joined
Feb 17, 2008
Messages
617
Location
In Space
Gender
Male
HSC
2009
blackglitter said:
Hey guys, can you please help me with the following questions. Thanks

1) Show that
(2sin3A + 2 cos3 A) / (sin A + cos A ) = 2 – sin2A
if sin A + cos A does not equal 0


2) [sin A / (cos A + sin A) ] + [sin A / (cos A - sin A) = tan2A

3) (2tanA) / (1 + tan2A) = sin 2A

4) (sin 3A/sinA) - (cos3A/cosA) = 2
When sin A and cos A do not equal 0

Thank you :)
Click to expand...

2) [sin A / (cos A + sin A) ] + [sin A / (cos A - sin A) = tan2A
LHS = [sin A (cos A - sin A) / cos2 A - sin2 A] + [sin A (cos A + sin A) / cos2 A - sin2 A]
=sinAcosA - sin2 A + sinAcosA + sin2 A / cos2 A - sin2 A
=2sinAcosA / cos2 A - sin2 A
=Sin2A/Cos2A
=Tan2A
=RHS
 
lyounamu

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
blackglitter said:
Hey guys, can you please help me with the following questions. Thanks

1) Show that
(2sin3A + 2 cos3 A) / (sin A + cos A ) = 2 – sin2A
if sin A + cos A does not equal 0


2) [sin A / (cos A + sin A) ] + [sin A / (cos A - sin A) = tan2A

3) (2tanA) / (1 + tan2A) = sin 2A

4) (sin 3A/sinA) - (cos3A/cosA) = 2
When sin A and cos A do not equal 0

Thank you :)
Click to expand...
2. sina/(cosa + sina) + sina/ ( cosa-sina)
sina(cosa-sina)/(cosa+sina)(cosa-sina) + sina(cosa+sina)/(cosa-sina)(cosa+sina)
= sinacosa - sin^2 + sinacosa + sin^2/cos^2-sin^2
= sin2a /cos2a
= tan 2a
blackglitter said:
Hey guys, can you please help me with the following questions. Thanks

1) Show that
(2sin3A + 2 cos3 A) / (sin A + cos A ) = 2 – sin2A
if sin A + cos A does not equal 0


2) [sin A / (cos A + sin A) ] + [sin A / (cos A - sin A) = tan2A

3) (2tanA) / (1 + tan2A) = sin 2A

4) (sin 3A/sinA) - (cos3A/cosA) = 2
When sin A and cos A do not equal 0

Thank you :)
Click to expand...
Q 3 is not right. It cannot be solved.

blackglitter said:
Hey guys, can you please help me with the following questions. Thanks

1) Show that
(2sin3A + 2 cos3 A) / (sin A + cos A ) = 2 – sin2A
if sin A + cos A does not equal 0


2) [sin A / (cos A + sin A) ] + [sin A / (cos A - sin A) = tan2A

3) (2tanA) / (1 + tan2A) = sin 2A

4) (sin 3A/sinA) - (cos3A/cosA) = 2
When sin A and cos A do not equal 0

Thank you :)
Click to expand...
4. shall do it later. Too tired of maths.
 
Last edited:
jules.09

jules.09

Member
Joined
Jun 23, 2007
Messages
360
Gender
Undisclosed
HSC
2009
4.

LHS
= (sin3A/sinA) - (cos3A/cosA)
= (sin3AcosA - cos3AsinA)/sinAcosA
= sin(3A-A)/sinAcosA
= sin2A/sinAcosA
= 2sin2A/2sinAcosA
= 2sin2A/sin2A
= 2
= RHS
 
jules.09

jules.09

Member
Joined
Jun 23, 2007
Messages
360
Gender
Undisclosed
HSC
2009
This is what I have so far for 1.

LHS
= (2sin3A+2cos3A)/(sinA+cosA)
= 2(sin3A+cos3A)/(sinA+cosA)
= 2[sin(2A+A)+cos(2A+A)]/(sinA+cosA)
= 2(sin2AcosA+cos2AsinA+cos2AcosA-sin2AsinA)/(sinA+cosA)
= 2[cos2A(sinA+cosA) - sin2A(sinA-cosA)]/(sinA+cosA)
= 2cos2A - 2sin2A[(sinA-cosA)/(sinA+cosA)]*[(sinA-cosA)/(sinA-cosA)]
= 2cos2A - 2sin2A[(sin²A+cos²A-2sinAcosA)/(sin²A-cos²A)]
= 2cos2A + 2sin2A[(1-sin2A)/cos2A)]
= 2cos2A + 2tan2A(1-sin2A)

... which is where namu got to towards the end. But I'll try to finish this off.
 
lyounamu

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
jules.09 said:
This is what I have so far for 1.

LHS
= (2sin3A+2cos3A)/(sinA+cosA)
= 2(sin3A+cos3A)/(sinA+cosA)
= 2[sin(2A+A)+cos(2A+A)]/(sinA+cosA)
= 2(sin2AcosA+cos2AsinA+cos2AcosA-sin2AsinA)/(sinA+cosA)
= 2[cos2A(sinA+cosA) - sin2A(sinA-cosA)]/(sinA+cosA)
= 2cos2A - 2sin2A[(sinA-cosA)/(sinA+cosA)]*[(sinA-cosA)/(sinA-cosA)]
= 2cos2A - 2sin2A[(sin²A+cos²A-2sinAcosA)/(sin²A-cos²A)]
= 2cos2A + 2sin2A[(1-sin2A)/cos2A)]
= 2cos2A + 2tan2A(1-sin2A)

... which is where namu got to towards the end. But I'll try to finish this off.
Click to expand...
But 1 is wrong. You cannot do it. (*just woke up from sleep*)
 
jules.09

jules.09

Member
Joined
Jun 23, 2007
Messages
360
Gender
Undisclosed
HSC
2009
lyounamu said:
But 1 is wrong. You cannot do it. (*just woke up from sleep*)
Click to expand...
Ok, I'll take your word that it's wrong. But how do you know it is wrong?

Is 3 also wrong in that case? :confused:
 
blackglitter

blackglitter

Member
Joined
May 25, 2008
Messages
198
Location
Somewhere on Earth
Gender
Female
HSC
2009
Thanks guys :) I couldn't do question 1 and that just threw me off the rest. I really hate dodgy questions...
 
Last edited:
M@ster P

M@ster P

Member
Joined
Feb 9, 2008
Messages
619
Gender
Male
HSC
2009
lyounamu said:
But 1 is wrong. You cannot do it. (*just woke up from sleep*)
Click to expand...
I think you can safely assume that the question is wrong since the working out that you did is like 3 pages long
 
lyounamu

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
Ok, I knew that q1 and 3 were un-doable because I put in random numbers into the equations and saw whether it worked or not. If you put, for example, 30 into all x into Q 1 and 3, you will see that they don't work out.
 
bored of sc

bored of sc

Active Member
Joined
Nov 10, 2007
Messages
2,314
Gender
Male
HSC
2009
blackglitter said:
Hey guys, can you please help me with the following questions. Thanks

1) Show that
(2sin3A + 2 cos3A) / (sin A + cos A ) = 2 – sin2A
if sin A + cos A does not equal 0
Click to expand...
Now it works.
 
ratcher0071

ratcher0071

Member
Joined
Feb 17, 2008
Messages
617
Location
In Space
Gender
Male
HSC
2009
1) Show that
(2sin3A + 2 cos3A) / (sin A + cos A ) = 2 – sin2A
if sin A + cos A does not equal 0

LHS=2(sin3A + cos3 A) / (sinA + cosA)
=2(sin2A + cos2 A)
=2(1)
=2
=/= RHS

:mad:
 
lyounamu

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
ratcher0071 said:
1) Show that
(2sin3A + 2 cos3A) / (sin A + cos A ) = 2 – sin2A
if sin A + cos A does not equal 0

LHS=2(sin3A + cos3 A) / (sinA + cosA)
=2(sin2A + cos2 A)
=2(1)
=2
=/= RHS

:mad:
Click to expand...
You made a mistake there. You cannot move directly from there to there. Hint: use a cubic rule.

2(sin^3(a) + cos^3(a))/(sina+cosa) = 2(sin^2(a) - sinacosa + cos^2(a)) (here I used a cubic rule that a^3 + b^3 = (a+b)(a^2-ab+b^2)
= 2-2sinaosa = 2-sin2a
 
bored of sc

bored of sc

Active Member
Joined
Nov 10, 2007
Messages
2,314
Gender
Male
HSC
2009
ratcher0071 said:
LHS=2(sin3A + cos3 A) / (sinA + cosA)
=2(sin2A + cos2 A)
=2(1)
=2
=/= RHS

:mad:
Click to expand...
LHS = 2(sin3A + cos3A) / (sin A + cos A)
= 2((sinA + cosA)(sin2A - sinAcosA + cos2A) / (sinA + cosA)
= 2(1 - sinAcosA)
= 2 - sin2A
= RHS
 
lyounamu

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
blackglitter, what about Q3? Can you give us the right question?

Will just assume that it is:

2tana/1+tan^2(a) = 2tana/sec^2(a) = 2tana . cos^2(a) = 2sinacosa = sin2a
 
You must log in or register to reply here.

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top