Math Doubts

Proof of Formula for subtracting Cos double angle from one

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.

Subtraction of cos double angle from one

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

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

Expand the cosine of double angle function

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

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

Simplify the trigonometric expression

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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