Let us take, $x$ is a variable that represents an angle of a right triangle and $2x$ is also an angle of another right angled triangle. In this case, the cosine of double angle function is written as $\cos{2x}$, and squares of sine and cosine functions are written as $\sin^2{x}$ and $\cos^2{x}$ respectively.

Now, let us derive a trigonometric identity that helps us to subtract the cosine of double angle function from one.

Write the subtraction of cosine of double angle function from one in mathematical form.

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

According to the cos double angle identity, expand the $\cos{2x}$ function in terms of sine and cosine of angle.

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

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

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

The square of cos function can be converted as square of sin function as per Pythagorean identity of sine and cosine functions.

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

Now, continue the simplification of the trigonometric expression.

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

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

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

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

Therefore, it is proved that the subtraction of cosine of double angle from one is equal to two times the square of sine of angle. If you want to derive the proof for a trigonometric expression either $1-\cos{2\theta}$ or $1-\cos{2A}$, then take $\theta$ or $A$ instead of $x$.

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