Finding the exact value of Tan[pi/8]

[suskieanna]

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 that to find the exact value of

Can anyone explain how I can approach this problem? Thanks in advance

[AlphaZero]

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 that to 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 \(\tan(\pi/8)\)

[lzxnl]

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 that to find the exact value of

Can anyone explain how I can approach this problem? Thanks in advance

Firstly, when you're using tex, type \tan(\frac{\pi}{4}), for instance.

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.