The word ‘trigonometry’ being driven from the Greek words’ ‘trigon’ and ‘metron’ and it means ‘measuring the sides of a triangle’. The subject originally thought and part of the scope of development to solve geometric problems involving triangles. We know about the trigonometric ratios of acute angles as the ratio of the sides of a right-angle triangle. In this Chapter, we will generalize the concept and Cos 2X formula of one such trigonometric ratios namely cos 2X with other trigonometric ratios. Let us start with our learning!

**Cos 2X Formula**

**What is a Cos 2X?**

The trigonometric ratios of an angle in a right triangle define the relationship between the angle and the length of its sides. Cosine 2X or Cos 2X is also, one such trigonometrical formula, also known as double angle formula, as it has a double angle in it.

Because of this, it is being driven by the expressions for trigonometric functions of the sum and difference of two numbers (angles) and related expressions.

Let us start with the cos two thetas or cos 2X or cosine of double angle formula.

**Derivation of Cos 2X Formula**

Let us start with the consideration of addition formula,

\(\cos \left ( X+Y \right )\) = \(\cos X \cos Y – \sin X \sin Y \)

Let us equate, X and Y, i.e. X = Y

So, the above formula for cos 2X, becomes

\(\cos 2X = \cos \left ( X+X \right ) = \cos X \cos X – \sin X \sin X \)

\(\cos 2X = \cos ^{2}X – \sin ^{2}X \)

Hence, the first cos 2X formula follows, as

\(\cos 2X = \cos ^{2}X – \sin ^{2}X\)

And for this reason, we know this formula as double the angle formula, because we are doubling the angle.

**Other Formulae of cos 2X**

**\(\cos 2X = 1 – 2 \sin ^{2}X \)**

To derive this, we need to start from the earlier derivation. As we already know that,

\(\cos 2X = \cos ^{2}X – \sin ^{2}X \)

\(\cos 2X = \left ( 1-\sin ^{2}X \right ) – \sin ^{2}X [Since, \cos^{2}X = \left ( 1-\sin ^{2}X \right )] \)

\(\cos 2X = 1 – \sin ^{2}X – \sin ^{2}X \)

So,

\(\cos 2X = 1 – \left (\sin ^{2}X + \sin ^{2}X \right) \)

\(Hence, \cos 2X = 1 – 2 \sin ^{2}X \)

**\(\cos 2X = 2 \cos ^{2}X – 1 \)**

To derive this, we need to start from the earlier derivation. As we already know that,

\(\cos 2X = \cos ^{2}X – \sin ^{2}X \)

\(\cos 2X = \cos ^{2}X – \left ( 1-\cos ^{2}X \right ) [Since, \sin^{2}X = \left ( 1-\cos ^{2}X \right ) \)

\(\cos 2X = \cos ^{2}X – 1 + \cos ^{2}X \)

\(\cos 2X = \left ( \cos ^{2}X + \cos ^{2}X \right ) – 1 \)

\(Hence, \cos 2X = 2\cos ^{2}X – 1 \)

**\(\cos 2X = \frac{1-\tan ^{2}X}{1+\tan ^{2}X} \)**

To derive this, we need to start from the earlier derivation. As we already know that,

\(\cos 2X = \cos ^{2}X – \sin ^{2}X \)

\(\cos 2X = \cos ^{2}X – \sin ^{2}X \)

\(\cos 2X = \frac{\cos ^{2}X – \sin ^{2}X}{1 } \)

\(\cos 2X = \frac{\cos ^{2}X – \sin ^{2}X}{\cos ^{2}X + \sin ^{2}X} [Since, cos ^{2}X + \sin ^{2}X = 1] \)

Dividing both numerator and denominator by \(\cos ^{2}\)X, we get

\(\cos 2X = \frac{1-\tan ^{2}X}{1+\tan ^{2}X} [ Since, \tan X = \frac{\sin X}{\cos X}] \)

Hence, \(\cos 2X = \frac{1-\tan ^{2}X}{1+\tan ^{2}X} \)

** ****Solved Examples**

Now that we have seen the formula of Cos 2X, let us try some examples to deepen our understanding.

**Q:** Prove that,

\( \cos 3X = 4 \cos ^{3}X – 3 \cos X, \)

Ans: We have,

\(\cos 3X = \cos \left ( 2X + X \right ) \)

\(\cos 3X = \cos 2X \cos X – \sin 2X \sin X \)

\(\cos 3X = \left ( 2\cos ^{2}X – 1 \right)\cos X – 2 \sin X\cos X\sin X\)

\(\cos 3X = \left ( 2\cos ^{2}X – 1 \right)\cos X – 2 \cos X\left ( 1-\cos ^{2}X \right )\)

\(\cos 3X = 2\cos ^{3}X – \cos X – 2 \cos X + 2 \cos ^{3}X\)

\(\cos 3X = 4\cos ^{3}X – 3 \cos X \)

[Hence, proved]

I get a different answer for first example.

I got Q1 as 20.5

median 23 and

Q3 26