Ratio of Double Angles- Urgent (1 Viewer)

[fuckit1991]

hey guys i was wondering if anyone could help me with a few triq questions:

1. Find tan (3A) and cos (3A) and hence show that tan (3A) -tan (A) = (2sin(A))/(cos(3A))

2. prove that sin^2 (5A) - sin^2 (3A) = sin(8A).sin(2A)

3. prove that (1+sin(2A))/(1-sin(2A)) = tan^2 (45 + A)

greatly appreciate any help

 

[Trebla]

hey guys i was wondering if anyone could help me with a few triq questions:

1. Find tan (3A) and cos (3A) and hence show that tan (3A) -tan (A) = (2sin(A))/(cos(3A))

2. prove that sin^2 (5A) - sin^2 (3A) = sin(8A).sin(2A)

3. prove that (1+sin(2A))/(1-sin(2A)) = tan^2 (45 + A)

greatly appreciate any help

For 3:
RHS = tan²(45 + A)
= [1 + tan A]² / [1 - tan A]²
= (1 + 2tan A + tan²A) / (1 - 2tan A + tan²A)
Multiply top and bottom by cos²A
= (cos²A + 2sin A cos A + sin²A) / (cos²A - 2sin A cos A + sin²A)
= (1 + 2sin A cos A) / (1 - 2sin A cos A)
= (1 + sin 2A) / (1 - sin 2A)
= LHS

 

[fuckit1991]

For 3:
RHS = tan²(45 + A)
= [1 + tan A]² / [1 - tan A]²
= (1 + 2tan A + tan²A) / (1 - 2tan A + tan²A)
Multiply top and bottom by cos²A
= (cos²A + 2sin A cos A + sin²A) / (cos²A - 2sin A cos A + sin²A)
= (1 + 2sin A cos A) / (1 - 2sin A cos A)
= (1 + sin 2A) / (1 - sin 2A)
= LHS

thanx a lot for your swift reply :uhhuh: trebla...i understood the answer except for how you got to the first line = [1 + tan A]² / [1 - tan A]²...if u cud explain that i wud appreciate it

 

[Timothy.Siu]

thanx a lot for your swift reply :uhhuh: trebla...i understood the answer except for how you got to the first line = [1 + tan A]² / [1 - tan A]²...if u cud explain that i wud appreciate it

he used compound angle rule for tan except with tan²2x which is fundamentally the same
tan(a+b)=(tana+tan b)/1-tan a tan b

 

[fuckit1991]

he used compound angle rule for tan except with tan²2x which is fundamentally the same
tan(a+b)=(tana+tan b)/1-tan a tan b

thanx for clearing that up i get it now :uhhuh:..its the basic identity squared and tan(45) =1...any takers for the other qs?

 

[Timothy.Siu]

hey guys i was wondering if anyone could help me with a few triq questions:

1. Find tan (3A) and cos (3A) and hence show that tan (3A) -tan (A) = (2sin(A))/(cos(3A))

2. prove that sin^2 (5A) - sin^2 (3A) = sin(8A).sin(2A)

3. prove that (1+sin(2A))/(1-sin(2A)) = tan^2 (45 + A)

greatly appreciate any help

2. LHS=(sin 5A+sin 3A)(sin 5A-sin3A)=(2sin 4A cos A)(2cos 4A sin A)=4(sin 4A cos 4A) (sin A cos A)=4(1/2sin 8A)(1/2 sin 2A)=sin (8A).sin (2A)

 

[fuckit1991]

2. LHS=(sin 5A+sin 3A)(sin 5A-sin3A)=(2sin 4A cos A)(2cos 4A sin A)=4(sin 4A cos 4A) (sin A cos A)=4(1/2sin 8A)(1/2 sin 2A)=sin (8A).sin (2A)

lol im such an idiot. i lost you completely in the first line of working:4(sin 4A cos 4A) (sin A cos A). could you please explain how you derived that...sorry for being such a pest

 

[Timothy.Siu]

lol im such an idiot. i lost you completely in the first line of working:4(sin 4A cos 4A) (sin A cos A). could you please explain how you derived that...sorry for being such a pest

nah ur not an idiot, and its fine, i'll try explain it

sin^2 (5A) - sin^2 (3A) = sin(8A).sin(2A)
so from the LHS, u can do difference of two squares,(sin 5A+sin 3A)(sin 5A-sin3A) and then theres a rule ( u mite not have learnt it) sin U + sin V=2sin (U+V)/2 x cos (U-V)/2
so i just used that to simplify that down to (2sin 4A cos A)(2cos 4A sin A) and from there u just multiply them and group together the 4A and A so u can use sin 2A=2sin Acos A to simplify it down and u'll get the answer

edit: go here http://community.boredofstudies.org/attachment.php?attachmentid=2976&d=1074163222
theres a thing on the method i used sin U + sin V=2sin (U+V)/2 x cos (U-V)/2
otherwise, u'd probably have to expand them and that wud take a long time

 

[fuckit1991]

nah ur not an idiot, and its fine, i'll try explain it

sin^2 (5A) - sin^2 (3A) = sin(8A).sin(2A)
so from the LHS, u can do difference of two squares,(sin 5A+sin 3A)(sin 5A-sin3A) and then theres a rule ( u mite not have learnt it) sin U + sin V=2sin (U+V)/2 x cos (U-V)/2
so i just used that to simplify that down to (2sin 4A cos A)(2cos 4A sin A) and from there u just multiply them and group together the 4A and A so u can use sin 2A=2sin Acos A to simplify it down and u'll get the answer

edit: go here http://community.boredofstudies.org/attachment.php?attachmentid=2976&d=1074163222
theres a thing on the method i used sin U + sin V=2sin (U+V)/2 x cos (U-V)/2
otherwise, u'd probably have to expand them and that wud take a long time

thanx a lot...i especially appreciated the link to the pdf of formulae. you must be a genius to get all this ext 1 stuff a whole year before ur hsc . from the identities in my text book there was no possible solution to this question...any takers for q1?

 

[fuckit1991]

btw timothy...do you know of any way to derive that sum and difference formula(sin U +Sin V...) from those identities given in the syllabus...hope im not taking too much of your time...greatly appreciate it

 

[Timothy.Siu]

btw timothy...do you know of any way to derive that sum and difference formula(sin U +Sin V...) from those identities given in the syllabus...hope im not taking too much of your time...greatly appreciate it

yeah

sin (A+B)=sin A cos B + sin B cos A
sin (A-B)=sin A cos B - sin B cos A
cos (A+B)=cos A cos B - sin A sin B
cos (A-B) = cos A cos B + sin A sin B

adding the first two equations, we get sin (A+B) + sin (A-B)= 2sin A cos B
basically...and u can do that for the other ones

 

[bored of sc]

hey guys i was wondering if anyone could help me with a few triq questions:

1. Find tan (3A) and cos (3A) and hence show that tan (3A) -tan (A) = (2sin(A))/(cos(3A))

tan3A
= tan(2A+A)
= (tan2A+tanA)/(1-tan2A*tanA)
= [(2tanA/1-tan²A)+tanA]/[1-(2tanA/1-tan²A)tanA]
= [(2tanA+tanA(1-tan²A))/(1-tan²A)]/[(1-tan²A-2tanA*tanA)/(1-tan²A)]
= (3tanA-tan³A)/(1-3tan²A)

cos3A
= cos(2A+A)
= cos2A*cosA - sin2A*sinA
= (cos²A-sin²A)cosA-2sin²AcosA
= cos³A-sin²AcosA-2sin²cosA
= cos³A-3sin²AcosA
= cos³A-3(1-cos²A)cosA
= 4cos³A-3

 

[Trebla]

hey guys i was wondering if anyone could help me with a few triq questions:

1. Find tan (3A) and cos (3A) and hence show that tan (3A) -tan (A) = (2sin(A))/(cos(3A))

2. prove that sin^2 (5A) - sin^2 (3A) = sin(8A).sin(2A)

3. prove that (1+sin(2A))/(1-sin(2A)) = tan^2 (45 + A)

greatly appreciate any help

For 1: (without using the hence...lol)
LHS = tan 3A - tan A
= sin 3A / cos 3A - sin A / cos A
= (sin 3A cos A - sin A cos 3A) / (cos 3A cos A)
= sin 2A / (cos 3A cos A)
= 2sin A cos A / (cos 3A cos A)
= 2sin A / cos 3A
= RHS

 

[bored of sc]

For 1: (without using the hence...lol)
LHS = tan 3A - tan A
= sin 3A / cos 3A - sin A / cos A
= (sin 3A cos A - sin A cos 3A) / (cos 3A cos A)
= sin 2A / (cos 3A cos A)
= 2sin A cos A / (cos 3A cos A)
= 2sin A / cos 3A
= RHS

Oh. It's that simple.