# Proof of Cos double angle identity in terms of Tan

Let us take, $x$ is a variable and also represents an angle of a right triangle. The sin, cos and tan functions are written as $\sin{x}$, $\cos{x}$ and $\tan{x}$ respectively. The cos of double angle function is written as $\cos{(2x)}$ in mathematics. Now, let us learn how to derive the cos double angle formula in terms of tan function.

### Expand the Cosine of double angle

Use the cosine of double angle identity for expanding the cosine of double angle function $\cos{2x}$ mathematically.

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

### Acceptable adjustment for getting required form

It is acceptable to express the trigonometric expression $\cos^2{x}-\sin^2{x}$ in ratio form and this trick helps us to obtain the required result in the upcoming steps.

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

The expression in the numerator is in terms of sine and cosine functions and let us try to express the denominator in the similar form. It is possible to express the number $1$ in terms of sine and cosine function by the Pythagorean identity of sin and cos functions.

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

### Steps for obtaining the Required result

It is possible to express the entire expression in terms of tan function only if the cosine function is brought under the sine function. So, let us try do it by a mathematical technique.

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

Express, the number $1$ as the quotient of the $\cos^2{x}$ by $\cos^2{x}$ functions.

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

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

$=\,\,\,$ $\dfrac{\cos^2{x}}{\cos^2{x}+\sin^2{x}}$ $\times$ $\Bigg(\dfrac{\cos^2{x}}{\cos^2{x}}-\dfrac{\sin^2{x}}{\cos^2{x}}\Bigg)$

$=\,\,\,$ $\dfrac{\cos^2{x}}{\cos^2{x}+\sin^2{x}}$ $\times$ $\require{cancel} \Bigg(\dfrac{\cancel{\cos^2{x}}}{\cancel{\cos^2{x}}}-\dfrac{\sin^2{x}}{\cos^2{x}}\Bigg)$

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

$=\,\,\,$ $\dfrac{\cos^2{x}}{\cos^2{x}+\sin^2{x}}$ $\times$ $\Bigg(1-\Big(\dfrac{\sin{x}}{\cos{x}}\Big)^2\Bigg)$

According the reciprocal identity of sine and cosine functions, the quotient of $\sin{x}$ by $\cos{x}$ is equal to $\tan{x}$.

$=\,\,\,$ $\dfrac{\cos^2{x}}{\cos^2{x}+\sin^2{x}}$ $\times$ $\Big(1-(\tan{x})^2\Big)$

$=\,\,\,$ $\dfrac{\cos^2{x}}{\cos^2{x}+\sin^2{x}}$ $\times$ $\Big(1-\tan^2{x}\Big)$

$=\,\,\,$ $\Big(1-\tan^2{x}\Big)$ $\times$ $\dfrac{\cos^2{x}}{\cos^2{x}+\sin^2{x}}$

$=\,\,\,$ $\Big(1-\tan^2{x}\Big)$ $\times$ $\dfrac{1 \times \cos^2{x}}{\cos^2{x}+\sin^2{x}}$

Now, express the second factor in its reciprocal form and then simplify the expression in the denominator.

$=\,\,\,$ $\Big(1-\tan^2{x}\Big)$ $\times$ $\dfrac{1}{\dfrac{\cos^2{x}+\sin^2{x}}{\cos^2{x}}}$

$=\,\,\,$ $\Big(1-\tan^2{x}\Big)$ $\times$ $\dfrac{1}{\Bigg(\dfrac{\cos^2{x}}{\cos^2{x}}+\dfrac{\sin^2{x}}{\cos^2{x}}\Bigg)}$

$=\,\,\,$ $\Big(1-\tan^2{x}\Big)$ $\times$ $\require{cancel} \dfrac{1}{\Bigg(\dfrac{\cancel{\cos^2{x}}}{\cancel{\cos^2{x}}}+\dfrac{\sin^2{x}}{\cos^2{x}}\Bigg)}$

$=\,\,\,$ $\Big(1-\tan^2{x}\Big)$ $\times$ $\dfrac{1}{\Bigg(1+\dfrac{\sin^2{x}}{\cos^2{x}}\Bigg)}$

$=\,\,\,$ $\Big(1-\tan^2{x}\Big)$ $\times$ $\dfrac{1}{\Big(1+\tan^2{x}\Big)}$

$=\,\,\,$ $\dfrac{\Big(1-\tan^2{x}\Big) \times 1}{\Big(1+\tan^2{x}\Big)}$

$=\,\,\,$ $\dfrac{\Big(1-\tan^2{x}\Big)}{\Big(1+\tan^2{x}\Big)}$

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

Therefore, it is proved that the cos of double angle is equal to the quotient of subtraction of square of tan function from one by the addition of one and square of tan function. If you want to derive the $\cos{(2A)}$ or $\cos{(2\theta)}$ rule in terms of tan function then take $A$ or $\theta$ respectively instated of $x$.

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