Question

help me recall a identity

Answer 1

how to find tanA if you know tan(A/2)

Answer 2

\[\cos^2(x)-\sin^2(x)=\cos(2x) \\ \cos^2(x)-(1-\cos^2(x))=\cos(2x) \\ \\ \text{ so we have } 2 \cos^2(x)-1=\cos(2x) \\ \text{ and } 1-2 \sin^2(x)=\cos(2x) \\ \text{ \let } x=\frac{A}{2} \\ \text{ so } 2x=A \\ \\ \text{so we have } 2 \cos^2(\frac{A}{2})-1=\cos(A) \\ \text{ and } 1-2 \sin^2(\frac{A}{2})=\sin(A) \]

Answer 3

\[\tan(A)=\frac{\sin(A)}{\cos(A)}=\frac{ 2 \cos^2(\frac{A}{2})-1}{1-2 \sin^2(\frac{A}{2})}\]
if I didn't make a mistake

Answer 4

thanks

Answer 5

\[\tan(A)=\frac{2-\sec^2(\frac{A}{2})}{\sec^2(\frac{A}{2})-2 \tan^2(\frac{A}{2})} \\ \tan(A)= \frac{2-(1+\tan^2(\frac{A}{2}))}{1+\tan^2(\frac{A}{2})-2\tan^2(\frac{A}{2})}\]
and I think you can simplify this to your liking

Answer 6

ok thank you =)

Answer 7

did i make a mistake there

Answer 8

I think I made mistake somewhere because that equals 1

Answer 9

oh I see what I did

Answer 10

I'm sorry

Answer 11

I have to start over

Answer 12

I accidentally called sin(A) something I shouldn't have :(

Answer 13

it's ok i get the idea , i will handle it , thanks

Answer 14

lol sorry
i feel dumb now

Answer 15

ok good luck @No.name
let me know if you get stuck and i will try again

Answer 16

yeah sinA , you made a mistake there the numerator is incorrect that is cosA itself lol

Answer 17

but fine i appreciate your help

Answer 18

right :p

Answer 19

that was really dumb on my part

Answer 20

I guess I just got cocky