inverse trig. question (1 Viewer)

[freaking_out]

hello, i think that this is a very simple question but i have a mind block at the moment :mad1:

1. prove that 2tan^-1(2)=pi - cos^-1(3/5)

 

[underthesun]

ah ha..

the equation equals

[ arctan = tan^-1, arccos = cos^-1 ok? ]

arctan(2) = pi/2 - arccos(3/5)/2

2 = tan((pi/2) - arccos(3/5)/2)

agree this far?

well now i'm having difficulty myself..

edit: after this i was gonna use double angle, but

how do you tan(arccos(3/5)/2)?

edit: ahha

let t = tan(theta/2)

cos(theta) = (1 - t^2)/(1+t^2)

shit..

isnt' tan(pi/2) infinite or something? undefined i guess

something wrong with the question, or something wrong with me?

 

[freaking_out]

well now i'm having difficulty myself..

that makes two of us

 

[underthesun]

see the edited post

where did you get this question again?

 

[wogboy]

prove that 2tan^-1(2)=pi - cos^-1(3/5)

I guess you could do it this way:

take cosines of both sides, so that:

cos(RHS)
= cos(pi - arccos[3/5])
= -cos(arccos[3/5])
= -3/5

(I prefer to write arccos rather than cos^-1 )

cos(LHS)
= cos(2*arctan[2])
= 2cos^2(arctan[2]) - 1

now cos^2(arctan[2]) = (1/sqrt[5])^2 = 1/5

(draw up the triangle to show this)
hence,

2cos^2(arctan[2]) - 1
= 2*(1/5) - 1
= 2/5 - 1
= -3/5

hence cos(LHS) = cos(RHS)

now since the range of the function arctan is 0 < arctan(x) < pi/2 for all positive real x, so we can say that:

0 < arctan[2] < pi/2
0 < 2*arctan[2] < pi

so,
0 < LHS < pi

also, we know that the range of the arccos function to be 0 < arccos(x) < pi, for all real x, so:

0 < arccos[3/5] < pi
-pi < -arccos[3/5] < 0
-pi + pi < pi - arccos[3/5] < 0 + pi
0< pi - arccos[3/5] < pi

so,
0 < RHS < pi

since cos(LHS) = cos(RHS) and 0 < LHS < pi and 0 < RHS < pi,

LHS = RHS

2arctan(2)=pi - arccos(3/5)

 

[freaking_out]

(I prefer to write arccos rather than cos^-1 )

:mad1:

 

[underthesun]

me too

that's a nice solution btw

 

[freaking_out]

Originally posted by wogboy
I guess you could do it this way:

take cosines of both sides, so that:

cos(RHS)
= cos(pi - arccos[3/5])
= -cos(arccos[3/5])
= -3/5

(I prefer to write arccos rather than cos^-1 )

cos(LHS)
= cos(2*arctan[2])
= 2cos^2(arctan[2]) - 1

now cos^2(arctan[2]) = (1/sqrt[5])^2 = 1/5

(draw up the triangle to show this)
hence,

2cos^2(arctan[2]) - 1
= 2*(1/5) - 1
= 2/5 - 1
= -3/5

hence cos(LHS) = cos(RHS)

now since the range of the function arctan is 0 < arctan(x) < pi/2 for all positive real x, so we can say that:

0 < arctan[2] < pi/2
0 < 2*arctan[2] < pi

so,
0 < LHS < pi

also, we know that the range of the arccos function to be 0 < arccos(x) < pi, for all real x, so:

0 < arccos[3/5] < pi
-pi < -arccos[3/5] < 0
-pi + pi < pi - arccos[3/5] < 0 + pi
0< pi - arccos[3/5] < pi

so,
0 < RHS < pi

since cos(LHS) = cos(RHS) and 0 < LHS < pi and 0 < RHS < pi,

LHS = RHS

2arctan(2)=pi - arccos(3/5)

i was just revising and i want to ask...is it really necessary to show how range of both sides so u can say that
cos(RHS)=cos(LHS)?

 

[wogboy]

i was just revising and i want to ask...is it really necessary to show how range of both sides so u can say that

Yes it certainly is. Just because cosA = cosB, it doesn't mean that A=B. This is because if you graph f(x)=cosx, you'll see that it goes up and down, and so one y-value corresponds to many x-values (at uni we say that such a function is NOT injective. Injective means that for each y value, there is no more than one corresponding x-value e.g. f(x)=x^3 is injective but f(x)=x^2 isn't. You don't need to worry about the terminology though).

However, f(x)=cosx is only decreasing between x=0 and x=pi, so you can say that it is injective for this small domain 0<=x<=pi. So if you can prove that cosA=cosB and also that A and B are both between 0 and pi, then it's good enough to say that A=B. Draw up a graph of f(x)=cosx, for 0<=x<=pi, and you'll see that it makes good sense.

 

[freaking_out]

oh ok, i never knew that, usually in exams i just say:

cos(LHS)=cos(RHS)

therefore: LHS=RHS [QED]

and my teacher didn't penalise marks for it either..

 

[wogboy]

and my teacher didn't penalise marks for it either..

Lucky you . My pedantic teacher back in high school would have lost me marks for doing that (she clearly made that point to us beforehand). But still it's better to play it safe and add in that range stuff in the HSC, in case you get a non generous marker.