Question

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

Answer 1

Find the value of $1-\sin \left(2x\right)$.

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 the equation $\left(2\right)$ into the given equation:

$1-\sin \left(2x\right)={\sin }^2\left(x\right)+{\cos }^2\left(x\right)-2\sin \left(x\right)\cos \left(x\right)...\left(3\right)$

Apply the algebraic identity ${\left(a-b\right)}^2=a^2+b^2-2ab$ in the equation $\left(3\right)$:

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

Hence, the value is $1-\sin \left(2x\right)={\left(\sin \left(x\right)-\cos \left(x\right)\right)}^2$.