Take, $x$ is a variable but represents an angle of a right triangle, $2x$ is a double angle but an angle of another right triangle. Therefore, the squares of sine and cosine functions are written as $\sin^2{x}$ and $\cos^2{x}$ respectively, and the cosine of double angle function is expressed as $\cos{2x}$ in mathematical form.

Now, let us start deriving the trigonometric identity that helps us to evaluate the sum of one and cosine of double angle function.

Add the cos double angle function to number one for expressing the addition in mathematical form.

$1+\cos{(2x)}$

As per the cos double angle identity, the $\cos{2x}$ function can be expanded in terms of sine and cosine of angle.

$\implies$ $1+\cos{(2x)}$ $\,=\,$ $1+(\cos^2{x}-\sin^2{x})$

Now, let’s simplify the trigonometric expression in the right-hand side of the trigonometric equation.

$=\,\,\,$ $1+\cos^2{x}-\sin^2{x}$

$=\,\,\,$ $1-\sin^2{x}+\cos^2{x}$

As per the Pythagorean identity of sine and cosine functions, the subtraction of sine squared of angle from one is equal to cosine squared of angle.

$=\,\,\,$ $\cos^2{x}+\cos^2{x}$

$\therefore \,\,\,\,\,\,$ $1+\cos{(2x)}$ $\,=\,$ $2\cos^2{x}$

Therefore, it is proved that the sum of one and cosine of double angle is equal to two times the cos squared of angle. If you want to express this proof for a trigonometric expression like either $1+\cos{2\theta}$ or $1+\cos{2A}$, then take $\theta$ or $A$ instead of $x$.

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