Inverse trig question (1 Viewer)

[Chris100]

Prove that 2tan^-12= Pi - cos^-13/5

Use the fact that tan(pi-x)=-tan x

 

[Demento1]

Did you let x = cos-1(3/5)?

 

[asianese]

Hint: Tan both sides independently.

 

[Chris100]

Asianese, I got tan(2tan^-12)=-tan(pi-cos^-13/5)
I don't know what to do next

 

[Chris100]

Nvm , above is wrong, I used the 4th quadrant.
This is where I'm at right now: tan(2tan^-12)=-tan(pi-cos^-1-3/5)

 

[xRyusenka]

Tan both sides so you are able to use the the given identity to evaluate the RHS.
Hint for LHS: let tan^-12 = a. So, tan(2a) = ?

 

[asianese]

Still having trouble?

 

[Chris100]

I can't tan the right hand side because it's out of the -pi/2 to pi/2 domain since we're dealing with inverse trig. Pi minus cos-1 3/5 is out of that domain
I'm probably spewing nonsense here though

 

[Chris100]

Lol nvm just got it thanks for the help guys.
Just to confirm though, in these sort of questions, do I not have to follow the traditional proof style of LHS=......=RHS?

 

[asianese]

Lol nvm just got it thanks for the help guys.
Just to confirm though, in these sort of questions, do I not have to follow the traditional proof style of LHS=......=RHS?

You dont have to but you must never work both sides simultaneously and end with 0=0 or something. You can do tan(lhs)=.... And then separately, tan(rhs)=.......=tan(lhs) and since lhs and rhs are acute, lhs=rhs