# Thread: sin cos equation

1. ## sin cos equation

can someone solve the equation
cos4x=sin2x in the easiest way possible using trigometric formula. i keep going around in circles and getting no where.

2. ## Re: sin cos equation

$\cos^{2}2x-\sin^{2}2x-\sin2x=0$
$1-\sin^{2}2x-\sin^{2}2x-\sin2x=0$
$2\sin^{2}2x+\sin2x-1=0,\ \sin2x=t$
$2t^2+t-1=0$
$\vdots$
$t_1=\frac12,\ t_2=-1$
$\ldots$

3. ## Re: sin cos equation

which formula converts cos4x into cos^2 2x -sin^2 2x

4. ## Re: sin cos equation

Originally Posted by markosheehan
which formula converts cos4x into cos^2 2x -sin^2 2x
The double angle formula for cosine: $\cos (2u) = \cos^2 u-\sin^2 u$. If $u = 2x$, you get $\cos(4x) = \cos^2(2x)-\sin^2(2x)$.

5. ## Re: sin cos equation

cos(4x)=sin(2x)
$1 - 2\sin^2(2x) = \sin(2x)$

$0 = 2\sin^2(2x) + \sin(2x) - 1$

$0 = [2\sin(2x)-1][\sin(2x)+1]$

$\sin(2x) = \dfrac{1}{2}$

$2x = \dfrac{\pi}{6} \pm 2k\pi \implies x = \dfrac{\pi}{12} \pm k\pi \text{ where } k \in \mathbb{Z}$

$2x = \dfrac{5\pi}{6} \pm 2k\pi \implies x = \dfrac{5\pi}{12} \pm k\pi \text{ where } k \in \mathbb{Z}$

$\sin(2x) = -1$

$2x = \dfrac{3\pi}{2} \pm k \cdot 2\pi \implies x = \dfrac{3\pi}{4} \pm k \cdot \pi \text{ where } k \in \mathbb{Z}$