Question

Find the value of the number C:

$\frac{1}{2}{\cos }^{2}x+C=\frac{1}{4}\cos \left(2x\right)$

Short answer

The value of C is $-\frac{1}{4}$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\frac{1}{2}{\cos }^{2}x+C=\frac{1}{4}\cos \left(2x\right)......\left(i\right)$

From the double-angle formulas,

$\cos \left(2\theta \right)=2{\cos }^2\theta -1$

From equation (i),

$$\begin{gathered} \frac{1}{2}{\cos }^{2}x+C=\frac{1}{4}\cos \left(2x\right) \\ C=\frac{1}{4}\cos \left(2x\right)-\frac{1}{2}{\cos }^{2}x \end{gathered}$$

Step 2. Simplify the equation.

We know $\cos 2x=2{\cos }^{2}x-1$, then

localid="1646414862206" $$\begin{gathered} C=\frac{1}{4}\left(2{\cos }^{2}x-1\right)-\frac{1}{2}c{\mathrm{os}}^{2}x \\ C=\frac{2}{4}{\cos }^{2}x-\frac{1}{4}-\frac{1}{2}c{\mathrm{os}}^{2}x \\ C=\frac{1}{2}{\cos }^{2}x-\frac{1}{4}-\frac{1}{2}c{\mathrm{os}}^{2}x \\ C=-\frac{1}{4} \end{gathered}$$