Question

How do you find the value of $\sin \left( 22\dfrac{1}{2} \right)$ using the double or half-angle formula?

Answer 1

We first try to assume the variables for the angle and the ratio $\sin \left( 22\dfrac{1}{2} \right)$. Then we use the theorem of sub-multiple angles where $\cos x=1-2{{\sin }^{2}}\dfrac{x}{2}$. We have to simplify the equation by using the binary operations. Then we take the root values and try to find the exact sign for the ratio.

Complete step-by-step solution:
We need to find the value of $\sin \left( 22\dfrac{1}{2} \right)$. We assume the variable $\sin \left( 22\dfrac{1}{2} \right)=m$
We know that the angle value of $\sin \left( 22\dfrac{1}{2} \right)$ is half of 45. So, ${{\left( \dfrac{45}{2} \right)}^{\circ }}={{\left( 22\dfrac{1}{2} \right)}^{\circ }}$.
We use the theorem of sub-multiple angles where $\cos x=1-2{{\sin }^{2}}\dfrac{x}{2}$.
To use the theorem, we are going to assume the value $x=45$ and we also need to find the value of $\cos \left( 45 \right)$.
We know that $\cos \left( 45 \right)=\dfrac{1}{\sqrt{2}}$.
The value of $\dfrac{x}{2}$ is $\dfrac{x}{2}=\left( \dfrac{45}{2} \right)=\left( 22\dfrac{1}{2} \right)$.
We now put these values to get
$\begin{aligned} & \cos x=1-2{{\left( \sin \dfrac{x}{2} \right)}^{2}} \\\ & \Rightarrow \dfrac{1}{\sqrt{2}}=1-2{{\left( m \right)}^{2}} \\\ \end{aligned}$
Simplifying the equation, we get $2{{m}^{2}}=1-\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}-1}{\sqrt{2}}$.
Now we need to find the value of m.
We can multiply $\sqrt{2}$ in the denominator and numerator to get ${{m}^{2}}=\dfrac{2-\sqrt{2}}{4}$.
Now we take the square root of both sides of the equation to get
$\begin{aligned} & \sqrt{{{m}^{2}}}=\pm \sqrt{\dfrac{2-\sqrt{2}}{4}} \\\ & \Rightarrow m=\pm \dfrac{\sqrt{2-\sqrt{2}}}{2} \\\ \end{aligned}$
We get two values for the ratio of $\sin \left( 22\dfrac{1}{2} \right)$. But it can’t be two values.
We know that $\sin 0=0$ and $\sin 90=1$. This means that the values of ratio sin in the interval of $\left[ 0,\dfrac{\pi }{2} \right]$ remain between 0 and 1.
So, the ratio of sin for angle $\sin \left( 22\dfrac{1}{2} \right)$ is in between 0 and 1 which means it will be positive.
So, $\sin \left( 22\dfrac{1}{2} \right)=m=\dfrac{\sqrt{2-\sqrt{2}}}{2}$.

Note: We can visualise the sin ratio from its graph value in the interval of $\left[ 0,\dfrac{\pi }{2} \right]$.

We can see that the graph for that interval is above the X-axis. So, even though the root value gives two signs we have to omit the negative value.