inverse trig help (1 Viewer)

[with-chu]

from y12 3u cambridge...

Prove:
2 inverse tan 2 = pi - inverse cos (3/5)

even with the hint that says "use the fact that tan(pi - x) = -tan" i can't seem to do it


help please!

 

[Trebla]

from y12 3u cambridge...

Prove:
2 inverse tan 2 = pi - inverse cos (3/5)

even with the hint that says "use the fact that tan(pi - x) = -tan" i can't seem to do it


help please!

Let a = tan^-12
=> tan a = 2

Let b = cos^-1(3/5)
=> 3/5 = cos b
=> tan b = 4/3

But
tan 2a = 2tan a / (1 - tan²a)
= - 4/3
= - tan b
= tan (π - b)
=> 2a = π - b
.: 2tan^-12 = π - cos^-1(3/5)

 

[Drongoski]

I will help you, if I'm able.

 

[with-chu]

Ah cool cool
big thanks to both posters!

 

[with-chu]

I have another question from the same exercise:

show that inv tan (1/2) + inv tan (2/5) + inv tan (8/9) = pi/2

only thing i can think of is letting each angle equal to alpha beta and gamma and taking tan of both sides but can't take the tan of pi/2... help please!

 

[Drongoski]

I have another question from the same exercise:

show that inv tan (1/2) + inv tan (2/5) + inv tan (8/9) = pi/2

only thing i can think of is letting each angle equal to alpha beta and gamma and taking tan of both sides but can't take the tan of pi/2... help please!

Actually when you get tan x = undefined ("infinity") then x = pi/2 (+- nxpi)
This annoying issue has cropped up again and again on bos.

In this case tan(a+b+c) = "infinity" ==> a+b+c = pi/2

or you can "avoid" this problem in my next post in LaTeX.

 

[Drongoski]

 

[with-chu]

got it now your solutions are so clear
thanks again!