Beyond Sins

$\cos4x = \cos2(2x)$ $= 1 - 2\sin^2(2x)$ $= 1 - 2(2\sin x\cdot\cos x)^2$ $1 - 8\sin^2x(1-\sin^2x)$ $= 1 - 8\sin^2x+8\sin^4x$ Putting the value: $= 1 - 32 + 128 = \fbox{97}$ .

It doesn't matter that our $x$ is not real; it is still true that $\sin^2 x + \cos^2 x = 1$ , and therefore, that all other known formulae still remain true.

Using this, we find $\cos x = \sqrt{-3}$ , a lovely imaginary number commonly represented by $\sqrt{3} \times i$ , where $i$ is the imaginary unit ( $i^2 = -1$ ).

With the help of double angle formulae, we can find the values of $\cos 2x$ and then $\cos 4x$ .

Let's start with $\sin 2x = 2 \sin x \cos x \Rightarrow \sin 2x = 2 \times 2 \times \sqrt{-3}\Rightarrow \sin 2x = 4 \sqrt{-3}$ .

And end with $\cos 4x =1 - 2 \sin^2 2x \Rightarrow \cos 4x = 1 - 2 \times 16 \times (-3) \Rightarrow\boxed{\cos 4x = 97}$ .

Isn't math beautiful? :)