Please help with year 12 trig

[pangb]

Express cot (A+B) in terms of cot A and cot B and hence find the exact value of cot 22.5. Deduce the exact value of cot 67.5.
Show that INVERSEcot(1+x) + INVERSEcot(1-x) = INVERSEcot((-x/2)^2)

I have got cot (A+B) = (cotAcotB -1)/(cotB+cotA)
then have no idea about the rest.

Any help would really be greatly appreciated, thanks a lot!

 

[tkhunny]

It's important to note that 45º = 22.5º + 22.5º and that 67.5º = 45º + 22.5º

 

[soroban]

Hello, pangb!

\(\displaystyle \text{Show that: }\:\cot^{\text{-}1}(1+x) + \cot^{\text{-}1}(1-x) \:=\: \cot^{\text{-}1}\!\left(\text{-}\dfrac{x^2}{2}\right)\)

Let \(\displaystyle A \,=\,\cot^{\text{-}1}(1+x) \quad\Rightarrow\quad \cot A \,=\,1+x\)
Let \(\displaystyle B \,=\,\cot^{\text{-}1}(1-x) \quad\Rightarrow\quad \cot B \,=\,1-x\)

The left side is: .\(\displaystyle A + B\)

Then:. \(\displaystyle \cot(A + B) \;=\;\dfrac{\cot A\cot B - 1}{\cot A + \cot B}\)

. . . . . \(\displaystyle \cot(A + B) \;=\; \dfrac{(1+x)(1-x) - 1}{(1+x) + (1-x)} \;=\; \)

. . . . . \(\displaystyle \cot(A + B) \;=\;\dfrac{1-x^2 - 1}{1+x+1-x} \;=\;\dfrac{\text{-}x^2}{2}\)


Therefore: .\(\displaystyle A + B \:=\:\cot^{\text{-}1}\!\left(\text{-}\dfrac{x^2}{2}\right)\)

 

[pangb]

You absolute hero, just in time as well.
Thank you very much soroban.