Question

How do you prove $\cos 2x = 2{\cos ^2}x - 1$ ?

Answer 1

Hint : In the given question, we have to verify that $\cos 2x = 2{\cos ^2}x - 1$ that is we have to prove that the left-hand side of the given equation is equal to the right-hand side, the left side of the equal to sign is called the left-hand side and the part on the right side of the equal to sign is called right-hand side. For that, we will take any one side and solve it using the trigonometric ratios or identities and make it equal to the other side.

Complete step-by-step answer :
Solving the left-hand side –
We know that $\cos 2x = {\cos ^2}x - {\sin ^2}x$
We also know that
${\sin ^2}x + {\cos ^2}x = 1 \\\ \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x \;$
So, we get –
$\cos 2x = {\cos ^2}x - (1 - {\cos ^2}x) \\\ \Rightarrow \cos 2x = {\cos ^2}x - 1 + {\cos ^2}x \\\ \Rightarrow \cos 2x = 2{\cos ^2}x - 1 \;$
The left-hand side has become equal to the right-hand side.
Hence proved that $\cos 2x = 2{\cos ^2}x - 1$ .
So, the correct answer is “$\cos 2x = 2{\cos ^2}x - 1$”.

Note : The trigonometric ratios are the ratio of two sides of a right-angled triangle. We know the value of the trigonometric functions of certain angles - $0,\dfrac{\pi }{6},\dfrac{\pi }{4},\dfrac{\pi }{3},\dfrac{\pi }{2}$ so to find out the trigonometric values of other angles, several trigonometric formulas have been made. One such formula states that the cosine of the sum of two angles is equal to the sum of the product of the cosine of each angle and the product of the sine of each angle, that is, $\cos (x + y) = \cos x\cos y - \sin x\sin y$
Putting y=x, in this formula, we get –
$\cos (x + x) = \cos x\cos x - \sin x\sin x \\\ \Rightarrow \cos 2x = {\cos ^2}x - {\sin ^2}x \;$
Trigonometry is a very vast field in mathematics and has applications in many of the other branches, so it becomes important to formulate such identities to make the calculations easier and efficient. We can solve the related questions using a similar approach.