# How do you prove cos²θ - sin²θ = 2cos²θ - 1?

Jul 5, 2016

see explanation.

#### Explanation:

To prove this identity we perform one of the following.

(1) Manipulate the left side until it equates to the right.

(2) Manipulate the right side until it equates to the left side.

(3) manipulate both sides until they equate to each other.

Here, using approach (1)

left side $= {\cos}^{2} \theta - {\sin}^{2} \theta \ldots \ldots . . \left(A\right)$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\sin}^{2} \theta + {\cos}^{2} \theta = 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

From this identity we can obtain.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\sin}^{2} \theta = 1 - {\cos}^{2} \theta} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Substitute this in (A)

$\Rightarrow {\cos}^{2} \theta - \left(1 - {\cos}^{2} \theta\right)$

$= {\cos}^{2} \theta - 1 + {\cos}^{2} \theta = 2 {\cos}^{2} \theta - 1 = \text{ right side}$

Thus we have proven that ${\cos}^{2} \theta - {\sin}^{2} \theta = 2 {\cos}^{2} \theta - 1$