Question

What is the value of $1+\sin \left(2x\right)$?

Answer 1

Determine the value of$\begin{array}{l}1+\sin \left(2x\right)\end{array}$.

Apply the trigonometric identities:

$\begin{array}{rcl}{\sin }^2\left(x\right)+{\cos }^2\left(x\right)& =& 1\cdots \left(1\right)\\ \sin \left(2x\right)& =& 2\sin \left(x\right)\cos \left(x\right)\cdots \left(2\right)\end{array}$

Substitute the value of $1$ from equation $\left(1\right)$ and the value of $\sin \left(2x\right)$ from equation $\left(2\right)$ in the given equation.

$$\begin{gathered} \begin{array}{rcl}1+\sin \left(2x\right)& =& {\sin }^2\left(x\right)+{\cos }^2\left(x\right)+2\sin \left(x\right)\cos \left(x\right)\end{array} \\ \begin{array}{rcl}& \Rightarrow & 1+\sin \left(2x\right)={\left(\sin \left(x\right)+\cos \left(x\right)\right)}^2[\because {\left(a+b\right)}^2=a^2+b^2+2ab]\end{array} \end{gathered}$$

Hence, the value is $\begin{array}{rcl}1+\sin \left(2x\right)& =& {\left(\sin \left(x\right)+\cos \left(x\right)\right)}^2\end{array}$.