Question

Find the value of the number C:

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

Short answer

The value ofCis $-\frac{1}{4}$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

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

Rearrange the equation,

role="math" localid="1646414063193" $C=-\frac{1}{4}\cos \left(2x\right)-\frac{1}{2}{\sin }^{2}x$

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

role="math" localid="1646414091146" $C=-\frac{1}{4}\left(1-2{\sin }^{2}x\right)-\frac{1}{2}{\sin }^{2}x$

Step 2. Simplify the equation.

Continuing the given equation,

$$\begin{gathered} C=-\frac{1}{4}+\frac{2}{4}{\sin }^{2}x-\frac{1}{2}{\sin }^{2}x \\ C=-\frac{1}{4}+\frac{1}{2}{\sin }^{2}x-\frac{1}{2}{\sin }^{2}x \\ C=-\frac{1}{4} \end{gathered}$$