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Question: How do you simplify \(2{{\cos }^{2}}\left( \dfrac{x}{2} \right)-1\) ?...

How do you simplify 2cos⁡2(x2)−12{{\cos }^{2}}\left( \dfrac{x}{2} \right)-1 ?

Explanation

Solution

We have been a trigonometric function, the cosine function which is in turn a function of a fractional value of x. In order to simplify this equation, we must apply basic trigonometric identities which have been derived and proven by application on the Pythagorean triangle. Then, we shall rearrange the equation formed and bring it to its simplest form.

Complete step by step answer:
While simplifying any trigonometric equation, we must always try to simplify them as the sine or cosine functions. This is because only these two functions are the simplest of all the trigonometric functions. Their values have been derived from the perpendicular, base and hypotenuse of a right-angled triangle according to the Pythagorean theory. All the other trigonometric functions such as tangent of angle, secant of angle, cotangent of angle and cosecant of angle are derived from these two basic functions only.
We must have a prior knowledge of basic trigonometric identities to solve this particular problem. We are given the square of the cosine function along with a constant term 1 being subtracted from it. Thus, we shall use the half angle identity,
2cos⁡2θ=cos⁡2θ+12{{\cos }^{2}}\theta =\cos 2\theta +1
⇒2cos⁡2θ−1=cos⁡2θ\Rightarrow 2{{\cos }^{2}}\theta -1=\cos 2\theta
In our given equation, θ=x2\theta =\dfrac{x}{2} and on applying the identity on the given equation, 2cos⁡2(x2)−12{{\cos }^{2}}\left( \dfrac{x}{2} \right)-1, we get:
⇒2cos⁡2(x2)−1=cos⁡2(x2)\Rightarrow 2{{\cos }^{2}}\left( \dfrac{x}{2} \right)-1=\cos 2\left( \dfrac{x}{2} \right)
⇒2cos⁡2(x2)−1=cos⁡x\Rightarrow 2{{\cos }^{2}}\left( \dfrac{x}{2} \right)-1=\cos x
Therefore, the given equation 2cos⁡2(x2)−12{{\cos }^{2}}\left( \dfrac{x}{2} \right)-1is simplified as equal to cos⁡x\cos x.

Note: We must take care that while simplifying trigonometric equations, all the half angles or square of angles and all other variations must be simplified to a simple linear angle in the given trigonometric function. Also, we must try to make the whole trigonometric equation with respect to any one trigonometric function only.