Cos2x Identity

Trigonometry, as it suggests, is all about triangles, mostly right-angled triangles. Trigonometry is a system that helps us to work out missing or unknown side lengths or angles in a triangle. Cos2x identity is one of the important identities in trigonometry that help us in finding the missing value.

In this article, we are going to look into cos2x identiy, its various formula, their derivations and work out some examples to grasp the concept in much better way.

History

The Greek mathematician Hipparchus produced the first known tables of trigonometry in about 140 BC. Although these tables have not survived, it is claimed that Hipparchus wrote twelve books of tables of chords, making Hipparchus the founder of trigonometry.

Important Cos2x Identity

Cos2x identity is one of the important identities in trigonometry that can be expressed in different ways. 

Cos2x identity can be expressed in terms of different trigonometric functions such as sine, cosine, and as well in tangent.

• \(\cos(2x) = \cos^2x – \sin^2x\)

• \(\cos(2x) = 2 \cos^2x – 1\)

• \(\cos(2x) = 1 – 2sin^2x\)

• \(\cos(2x) = \frac{1 – tan^2x}{1 + tan^2x}\)

Derivations of Cos2x Identity

1. \(\cos(2x) = \cos^2x – \sin^2x\)

\(
\begin{array}
\,\,\,\,\,\,\,\, \cos(2x) & = \cos(x + x)\\
\,\,\,\,\,\,\,\, & = \cos(x) \cos(x) – \sin(x) \sin(x)\\
\,\,\,\,\,\,\,\, & = \cos^2x – \sin^2x\\
\end{array}
\)

2. \(cos(2x) =  2 \cos^2x – 1\)

\(
\begin{array}
\,\,\,\,\,\,\,\, \cos(2x) & = \cos^2x – \sin^2x\\
\,\,\,\,\,\,\,\, & = \cos^2x – {1 – \cos^2x}\\
\,\,\,\,\,\,\,\, & = 2 \cos^2x – 1\\
\end{array}
\)

3. \(\cos(2x) = 1 – 2 \sin^2x\)

\(
\begin{array}
\,\,\,\,\,\,\,\, \cos(2x) & = \cos^2x – \sin^2x\\
\,\,\,\,\,\,\,\, & = {1 – \sin^2x} – \sin^2x\\
\,\,\,\,\,\,\,\, & = 1 – 2 \sin^2x\\
\end{array}
\)

4. \(\cos(2x) = \frac{1 – \tan^2x}{1 + \tan^2x}\)

\(
\begin{array}
\,\,\,\,\,\,\,\, \cos(2x) & = \cos^2x – \sin^2x\\
\,\,\,\,\,\,\,\, & = \frac{\cos^2x – \sin^2x}{1}\\
\,\,\,\,\,\,\,\, & = \frac{\cos^2x – \sin^2x}{\cos^2x + \sin^2x}\\
\end{array}\\
\,\,\,\,\,\,\,\, \text{Dividing numerator and denominator with} \cos^2x \\
\begin{array}\\
\,\,\,\,\,\,\,\, \cos(2x) & =\frac{ 1 – \frac{\sin^2x}{\cos^2x}}{1 + \frac{\sin^2x}{\cos^2x}} \\
\,\,\,\,\,\,\,\, & = \frac{1 – \tan^2x}{1 + \tan^2x}\\
\end{array}
\)

Solved Questions

Question 1. Find the value of \(\cos(120)\).

Solution.

\(
\begin{array}
\,\,\,\,\,\,\,\, \cos(120) & = \cos(2(60))\\
\,\,\,\,\,\,\,\, & = \cos^2(60) – \sin^2(60)\\
\,\,\,\,\,\,\,\, & = \frac{1}{2} – \frac{{3}}{4}\\
\,\,\,\,\,\,\,\, & = \frac{-1}{2}\\
\end{array}
\)

Question 2. Solve \(\cos(2a) = \sin(a)\), for \(-\pi \leq a \leq \pi.\)

Solution. 

\(\cos 2a = \sin a\)

\(1 – 2 \sin^2(a) = \sin a\)

\(2 \sin^2(a) + \sin(a) + 1 = 0\)

\((2 \sin(a) – 1)(\sin(a) + 1) = 0\)

Therefore, \(2 \sin(a) = 1\) or \(\sin(a) = (-1)\)

\(\sin(a) = \frac{1}{2}\) (or) \((-1)\)

FAQs

Related Articles

1. Sin2x Identity
2. Tan2x Identity