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Further Trig maths help (1 Viewer)

blackglitter

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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

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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

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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 !!!!
 
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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

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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

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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

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4.

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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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
 
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