Question

How do you find the exact value of $\cos 2x$ using the double angle formula?

Answer 1

We know that the above-given equation contains trigonometric functions, as the term cosine or $\cos$is a basic trigonometric ratio. Therefore we will use the trigonometric identities to get the solution of this question. Trigonometric equations that have multiple angle terms like as given in the above equation can be simplified using trigonometric identities.

Complete step-by-step answer:
Here we have an equation $\cos 2x$.
In this expression we have a double angle so we have to apply the double angle formula.
We know the identity that
$\cos (a + b) = \cos a \times \cos b + \cos b \times \cos a$
We can also write the given equation as sum of two angles:
$\cos (x + x)$
Therefore by applying the above trigonometric identity in the expression we can write:
$= \cos x \times \cos x - \sin x \times \sin x$
It gives us value
$\Rightarrow \cos 2x = {\cos ^2}x - {\sin ^2}x$
Now again by the above-derived formula, we can write
$\Rightarrow \cos 2x = {\cos ^2}x - {\sin ^2}x$
And we can write
$\Rightarrow {\sin ^2}x = 1 - {\cos ^2}x$
Therefore by applying this we can write
$= {\cos ^2}x - (1 - {\sin ^2}x)$4
On breaking the bracket and adding the terms we get:
$= {\cos ^2}x - 1 + {\cos ^2}x$
It gives us another new formula i.e.
$= 2{\cos ^2}x - 1$
Similarly using the first formula we can write
$\Rightarrow {\cos ^2}x - {\sin ^2}x = (1 - {\sin ^2}x) - {\sin ^2}x$
On breaking the brackets and arranging the terms we have:
$= 1 - 2{\sin ^2}x$
Hence we have received three new formulas of $\cos 2x$ using the double angle formula.

Note: We should know that the double angle formula is a trigonometric identity that expresses a trigonometric function of $2\theta$ in the term of a trigonometric function of $\theta$ . We should also note that the remember the general solutions of trigonometric functions before solving the sums as the general solution of cosine function at $0$ is $(n\pi + \dfrac{\pi }{2})$, also when cosine function is $1$ we have $x = 2n\pi$.