# Question #8e0f7

Jul 5, 2016

See the Proof in Explanation.

#### Explanation:

We use the Formula $: \cos \left(A + B\right) = \cos A \cos B - \sin A S \in B .$

Letting $A = B = x$, we get,

$\cos \left(x + x\right) = \cos x \cdot \cos x - \sin x \cdot \sin x$

$\therefore \cos 2 x = {\cos}^{2} x - {\sin}^{2} x ,$ or, ${\sin}^{2} x + \cos 2 x = {\cos}^{2} x .$

Hence, the Proof.

Jul 5, 2016

See below.

#### Explanation:

Answering this question requires the use of two important identities:

• ${\sin}^{2} x + {\cos}^{2} x = 1 \to$ Pythagorean Identity
• $\cos 2 x = {\cos}^{2} x - {\sin}^{2} x \to$ Double angle identity for cosine

Note that subtracting ${\cos}^{2} x$ from both sides in the first identity yields ${\sin}^{2} x = 1 - {\cos}^{2} x$, and it's this modified form of the Pythagorean Identity we will be using.

Now that we have a few identities to work with, we can do some substituting in ${\sin}^{2} x + \cos 2 x = {\cos}^{2} x$:
$\underbrace{1 - {\cos}^{2} x} + \underbrace{{\cos}^{2} x - {\sin}^{2} x} = {\cos}^{2} x$
$\textcolor{w h i t e}{X} {\sin}^{2} x \textcolor{w h i t e}{X X X X X} \cos 2 x$

We see that the cosines cancel:
$1 - \cancel{{\cos}^{2} x} + \cancel{{\cos}^{2} x} - {\sin}^{2} x = {\cos}^{2} x$
$\to 1 - {\sin}^{2} x = {\cos}^{2} x$

This is another form of the Pythagorean Identity ${\sin}^{2} x + {\cos}^{2} x = 1$; see what happens you you subtract ${\sin}^{2} x$ from both sides:
${\sin}^{2} x + {\cos}^{2} x = 1$
${\sin}^{2} x + {\cos}^{2} x - {\sin}^{2} x = 1 - {\sin}^{2} x$
$\cancel{{\sin}^{2} x} + {\cos}^{2} x - \cancel{{\sin}^{2} x} = 1 - {\sin}^{2} x$
$\to {\cos}^{2} x = 1 - {\sin}^{2} x$

That's exactly what we have in $1 - {\sin}^{2} x = {\cos}^{2} x$, so we can complete the proof:
${\cos}^{2} x = {\cos}^{2} x$