Easy Inverse Trig Problem. Help please! (1 Viewer)

[foram]

I can't do this (I know I may sound stupid now, but everybody has their dumb moments). Can sombody show me the proof for this please?

Prove:

tan^-1 (1/2) + tan^-1 (1/3) = pi/4

 

[conics2008]

this is not an inverse problem. Its a complex number problem... wait let me do it on paper first

 

[foram]

this is not an inverse problem. Its a complex number problem... wait let me do it on paper first

But this is in the inverse section of my understanding yr 12 math textbook... Is there some 3U method to prove this?

 

[conics2008]

not sure buddy, we are starting inverse this week.. so i dont know yet, its a short working out if you do it 4unit style lol....

 

[foram]

not sure buddy, we are starting inverse this week.. so i dont know yet, its a short working out if you do it 4unit style lol....

can you show me please?

 

[conics2008]

ok but first you need to understand complex numbers...

ok i will give you this rule if you're addin arguments ( angels ), its the same as multiplin complex numbers..

This is how it all starts.. argument of tan^-1 1/2 ==> tan(x)=y/x this comes from z=x+iy

now your y =1 and your x =2 therefore the complex number becomes

2+1i

now with your other argument tan^-1 (1/3) y=1 x=3

therefore complex number becomes 3+1i

since it says add you must multiply those two complex numbers, but in a case where its minus you divide your complex number.

any who, (3+i)(2+i) = 5+5i keep in mind i^2 = -1

now find your argument which is basically y/x which is tan(x)=5/5 > 1

where tan^-1 1 = 45 or pi/4

I know i never make sense here on BOS, but in writing i make alot more sense. I hope someone can come here and explain it to you .. I'm sorry.

 

[foram]

Actually, I just managed to solve it using 3U method.

let tan^-1 (1/2) = x

therefore tan x = 1/2

let tan^-1 (1/3) = y

therefore tan y = 1/3

tan (x + y) = (tan x + tan y) / (1 - tan x . tan y)
tan (x + y) = (1/2 + 1/3) / ( 1 - 1/2 - 1/3)
tan (x + y) = (5/6) / (5/6)
tan (x + y) = 1

x + y = tan^-1 (1)
x + y = pi/4

, I can't believe I didn't see something so simple! I'm still curious about the 4U method though.

 

[conics2008]

nice .. expansion style.. thats old school..

I would of just went with my 4unit style, because in the 4unit cambridge text book it has question like this I think... I remember doing question like them.. but non came in the test..


this question is basically arguments and moduls problems.. its not that big..

which book did you get this from ?

 

[foram]

understanding yr 12 maths Extention 1

 

[conics2008]

understanding yr 12 maths Extention 1

wow never heard of those text books, the only books i've heard are fitzpatrick, cambridge ( the ones i use ) and MATHS IN FOCUS my favourite .... lol

 

[foram]

wow never heard of those text books, the only books i've heard are fitzpatrick, cambridge ( the ones i use ) and MATHS IN FOCUS my favourite .... lol

Understanding series is a short, revision/cramming, textbook. The questions arn't very hard either, proberbly harder than MIF though. Everything is harder than MIF.

 

[barry1]

OMG i have that book as well (the one that has a light pink colour xD)
good to see i'm not alone =D, I reckon you need that book with a cambridge book to essentially kick any question that comes flying towards you once you've studied the topic
Good work!

 

[conics2008]

xD.. maths in focus makes me laugh lol...

Its full of crap, i see no help from it, but i do use this topic by topic book, its helpful, it gives you exam styles question with full working out...

cambridge is the shittest at working out, they hardly give you the dam answers..
Fitzpatrick is kind of lost... i dont know this is how i see them..

 

[foram]

I have another question I can't do.

Show that:

2tan^-1 (x) = tan^-1 [(2x)/(1 - x^2)] ;Provided that -1 < x < 1

 

[leoyh]

RHS = tan^-1 [ 2x / (1-x^2) ]

Let that = tan (A+B)

tan (A+B) = tan A + tan B / (1- tanAtanB)

= x + x / (1- x times x), by just reading and equating your tan (A+B) and the inverse tan

therefore tan A = x, tan B = x

i.e. tan^-1 (x) = A, tan^-1 (x) = B

A + B = tan^-1 (x) + tan^-1 (x) = 2tan^-1(x) = LHS

therefore, LHS = RHS

thats the easiest way i see of doing it. theres probably some better way but i've forgotten most of my inverse trig so yeah

hope it helps

 

[foram]

thankyou