cos2x +sin2x

I'll need to tướng memorize $\cos2x = \cos^2x - \sin^2x$ as I'll use it in derivatives.

Only, there are other forms for this identity, I can't see how I can get to tướng the others from this one above.

The other forms are

$$2\cos^2x - 1\\ 1 - 2\sin^2x$$

You can bởi it by using the Pythagorean identity: $\sin^2 x+\cos^2 x =1$. This can be rewritten two different ways:

$$\sin^2 x = 1- \cos^2 x$$

and

$$\cos^2 x = 1 - \sin^2 x$$

Use either of these formulas to tướng replace the $\sin^2 x$, or the $\cos^2 x$, on the right side of your identity. That will give you the other two forms.

answered Jul 30, 2017 at 1:02

answered Jul 30, 2017 at 1:04

This is because $\sin^2x + \cos^2=1$

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

answered Jul 30, 2017 at 1:01

I remember this easiest using Euler's identity:

$$ e^{i2x} = \cos2x + i\sin2x $$

and remembering that $e^{inx} = \left(e^{ix}\right)^n$ consider,

$$ \left(e^{ix}\right)^2 = \left(\cos x + i\sin x \right)^2 = \cos^2 x + 2i\cos x\sin x - \sin^2 x $$

Now equate the real parts, $$\Re[e^{i2x}] = \Re[\left(e^{ix}\right)^2]\implies \cos2x=\cos^2x-\sin^2x $$

answered Jul 30, 2017 at 1:16

\begin{align} \cos 2\theta &= \cos^2 \theta - \sin^2 \theta\\ 1 &= \cos^2 \theta + \sin^2 \theta \\ \end{align}

Adding, you get \begin{align} \cos 2\theta + 1 &= 2 \cos^2 \theta \\ \cos 2\theta &= 2 \cos^2 \theta - 1 \\ \end{align}

Subtracting, you get \begin{align} \cos 2\theta - 1 &= -2 \sin^2 \theta \\ \cos 2\theta &= 1 - 2 \cos^2 \theta \\ \end{align}

answered Jul 30, 2017 at 5:15