Question

if α and β are acute angles and cos2α=3cos2β-1/3-cos2β , show that tanα=√2 tanβ

Answer 1

3-cos2β is the denominator for 3cos2β too....

Answer 2

This is just manipulation of trigonometric identities.

Answer 3

\[cos (2\alpha )= \dfrac{3cos (2\beta) -1}{3-cos(2\beta)}\]

Answer 4

^^ Exactly

Answer 5

wow, it's tough man!!

Answer 6

I know.... :/

Answer 7

@wio help!

Answer 8

Surrender!! I don't know how to prove. I am sorry for wasting your time

Answer 9

the thing you get to assume has double angles while think you want to show doesn't
i think this is our starting point

Answer 10

@myininaya we have formula link them together. It is \(\huge tan\alpha =\dfrac{1-cos2\alpha}{sin2\alpha}\) and I replace cos 2 alpha but I got nothing :(

Answer 11

I tried many ways, even thought of to figure what it is. Nothing rings to me. :(

Answer 12

Right now what I'm doing is playing with it.
If I come up with something I will give you a hint in that direction.
Don't stop trying though. Just play and try to have fun. It is like a wonderful puzzle.

Answer 13

i have something you are still here

Answer 14

your triangle idea actually gave it to me

Answer 15

except kinda

Answer 16

so we are given \[\cos(2 \alpha)=\frac{3\cos(2 \beta)-1}{3-\cos(2 \beta)}\]
I built a triangle with 2 alpha as a measurement for a right triangle.
I use the top of that above ratio as the measurement for adjacent side of that triangle
and the bottom of that ratio be the hyp measurement for that triangle

Answer 17

then I found the opp side in terms of those measurements

Answer 18

then you said \[\tan(\alpha)=\frac{1-\cos(2 \alpha)}{\sin(2 \alpha)}\]
so I plug in what was given for cos(2 alpha) and what i found for sin(2 alpha)

Answer 19

then get rid of the compound fraction

Answer 20

and you will start to see something beautiful

Answer 21

like it is more than beautiful is actually what you want to obtain
you will have to use the fact that \[\tan(\beta)=\frac{1-\cos(2 \beta)}{\sin(2 \beta)}\]
at the very end there

Answer 22

ok well if you return let me know if you tried this and if you got stuck somewhere...

Answer 23

@myininaya
I am pretty close to getting it, but trying to find my arithmetical error.
the denominator seems to reduce to cos^2(b), and the numerator seems to be edging close to cos^2(b)-2sin^2(b).
I'll write it up and see if it makes sense.

Answer 24

Oops. Sorry @bebong I messaged you when this was actually @bajji 's problem. My bad!

Answer 25

Did you do my approach or another approach.

Answer 26

@myininaya I just blindly expand everything and hope to put back together something.
lol :)

Answer 27

given

then I found opposite side:
\[\text{ opp side measurement of } (2 \alpha)=\sqrt{(3-\cos(2 \beta))^2-(3\cos(2 \beta)-1)^2} \\ =\sqrt{(9-6\cos(2 \beta)+\cos^2(2 \beta))-(9\cos^2(2 \beta)-6\cos(2 \beta)+1) } \\ = \sqrt{8-8\cos^2(2 \beta)}=2 \sqrt{2} \sqrt{1-\cos^2(2 \beta)}\]
Just a little more detail about the direction I took
See how lovely that looks.