Post by: suskieanna on February 06, 2019, 11:47:00 am
Use the double angle formula for Tan[2x] and the fact that
Can anyone explain how I can approach this problem? Thanks in advance
Post by: AlphaZero on February 06, 2019, 03:53:07 pm
Hello I am trying to work on some textbook questions and I am a bit stuck :'( The question is:
Use the double angle formula for Tan[2x] and the fact thatto find the exact value of
Can anyone explain how I can approach this problem? Thanks in advance
Hey there.
From the formula sheet, \(\tan(2\theta)=\dfrac{2\tan(\theta)}{1-\tan^2(\theta)}\).
Substituting \(\theta=\dfrac\pi8\) gives us \(1=\dfrac{2\tan(\pi/8)}{1-\tan^2(\pi/8)},\) which after some rearranging, is just a quadratic equation in \(\tan(\pi/8)\) :)
Post by: lzxnl on February 06, 2019, 04:42:55 pm
Hello I am trying to work on some textbook questions and I am a bit stuck :'( The question is:Firstly, when you're using tex, type \tan(\frac{\pi}{4}), for instance.
Use the double angle formula for Tan[2x] and the fact that
Can anyone explain how I can approach this problem? Thanks in advance
Here is a little-known shortcut that solves these tangent half angle problems immediately. You can only use this method if the question doesn't force you to use the tangent double angle formula.
We wish to write \(\tan\left(x\right)\) in terms of \(\sin(2x),\cos(2x)\). Note that \(\sin(2x) = 2\sin(x)\cos(x),\cos(2x) = 2\cos^2(x) - 1 = 1 - 2\sin^2(x)\). We will need these.
We want to convert the \(\sin(x)\) and the \(\cos(x)\) into \(\sin(2x)\) etc. Can you see that we can multiply \(\cos(x)\) by \(2\sin(x)\) to get \(\sin(2x)\)? Multiply top and bottom by \(2\sin(x)\) to get
Now sub in \(x = \frac{\pi}{8}\) and you're done. No quadratics needed.