Simplifying this expression, trigonometry

Question

I have been having trouble understanding how $$6-6\cos\left(\frac{\pi}{4}\right) = 3\sqrt{2}.$$ My main problem is the conversion of the two separate terms into one.

Answer 1

Here's your triangle.

It's a right triangle and it's an isosceles triangle as well. We know it's an isosceles triangle, because the angles have to add up to $\pi$. Given that it has a right angle ($\pi/2$) and one of the angles is $\pi/4$, the other angle has to be $\pi/4$ as well. So, isosceles it is, with the two legs equal.

So, by Pythagoras, $$a^2+a^2=c^2\implies 2a^2=c^2 \implies \frac{a}{c}=\frac{1}{\sqrt 2} = \frac {\sqrt 2}{2}$$

But: $$\cos \frac {\pi}{4} = \frac{a}{c}= \frac {\sqrt 2}{2}$$

Plug that into your original expression:

$$6-6\cos\frac{\pi}{4}= 6 - 6(\frac {\sqrt 2}{2})=6 - 3\sqrt 2$$

So, the second term is equal to $3\sqrt 2$, rather than the whole expression.