Question

How do you verify ${{\cos }^{2}}2A=\dfrac{1+\cos 4A}{2}$ ?

Answer 1

Hint: First analyse the left hand side of the equation and check whether it is equal to the right hand side of the equation. This can be done by using the suitable trigonometric identities (formulae) and simplify any one side of the equation.

Complete step by step solution:
Sine, cosine and tangent of an angle are trigonometric ratios. These ratios are also called trigonometric functions. When we plot a graph of the trigonometric ratios with respect to all the real values of an angle, we get a graph that has a periodic property. This means that the graph repeats itself after equal intervals of the angle. Other than the trigonometric ratios sine, cosine and tangent we have other trigonometric ratios called cosecant, secant and cotangent.

All the above six trigonometric ratios (functions) are dependent on each other. There are different properties and identities that relate the trigonometric ratios. The equation that has to be verified is,
${{\cos }^{2}}2A=\dfrac{1+\cos 4A}{2}$
To verify the given equation we shall use the identity $\cos (x+y)=\cos x\cos x-\sin x\sin y$.
In this, substitute $x=y=2A$
Then, we get that
$\cos (2A+2A)=\cos 2A\cos 2A-\sin 2A\sin 2A$
$\Rightarrow \cos (4A)={{\cos }^{2}}2A-{{\sin }^{2}}2A$ …. (i)

We know that ${{\sin }^{2}}2A=1-{{\cos }^{2}}2A$
Substitute this value in (i)
$\cos (4A)={{\cos }^{2}}2A-\left( 1-{{\cos }^{2}}2A \right)$
$\Rightarrow \cos (4A)={{\cos }^{2}}2A-1+{{\cos }^{2}}2A$
Then,
$\therefore\cos (4A)=2{{\cos }^{2}}2A-1$
This gives us that ${{\cos }^{2}}2A=\dfrac{1+\cos 4A}{2}$.

Therefore, the given equation is correct.

Note:You can directly verify the given equation if you know the identity which says that $\cos 2x=1+2{{\cos }^{2}}x$. From this identity we can write that,
${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}$
Now, substitute $x=2A$ in the above equation.
Then we get ${{\cos }^{2}}2A=\dfrac{1+\cos 4A}{2}$