Question

How do you simplify using the half angle formula: - $\sin \left( {{22}^{\circ }},30' \right)$?

Answer 1

First of all convert 30’ into degrees by using the conversion relation: - $1'=\dfrac{1}{60}$ degrees. Using this conversion, find the total angle, i.e., ${{22}^{\circ }}30'$ in degrees. Assume this angle as $\theta$. Find the value of $2\theta$ by multiplying $\theta$ with 2. Now, use the trigonometric identity: - ${{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}$ and take square root both the sides to get the answer. Use the value: - $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$.

Complete step by step answer:
Here, we have been provided with the trigonometric function: - $\sin \left( {{22}^{\circ }},30' \right)$ and we are asked to simplify it using the half angle formula.
Now, first we need to convert the given angle into degrees. The angle is ${{22}^{\circ }}30'$ which is read as 22 degrees 30 minutes. We know that $60'={{1}^{\circ }}$, therefore we have,

& \Rightarrow 1'={{\left( \dfrac{1}{60} \right)}^{\circ }} \\\ & \Rightarrow 30'={{\left( \dfrac{30}{60} \right)}^{\circ }} \\\ & \Rightarrow 30'={{0.5}^{\circ }} \\\ \end{aligned}$$ So, the given angle is $${{22.5}^{\circ }}$$. Let us assume this angle as $$\theta $$. So, multiplying both the sides with 2, we get, $$\begin{aligned} & \Rightarrow 2\times \theta =2\times {{22.5}^{\circ }} \\\ & \Rightarrow 2\theta ={{45}^{\circ }} \\\ \end{aligned}$$ Using the trigonometric identity: - $${{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}$$, we get, $$\Rightarrow {{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}$$ Substituting the value of $$2\theta $$, we get, $$\Rightarrow {{\sin }^{2}}\theta =\dfrac{1-\cos {{45}^{\circ }}}{2}$$ Using the value: - $$\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$$, we get, $$\Rightarrow {{\sin }^{2}}\theta =\dfrac{1-\dfrac{1}{\sqrt{2}}}{2}=\dfrac{\sqrt{2}-1}{2\sqrt{2}}$$ Taking square root both the sides, we get, $$\Rightarrow \sin \theta ={{\left( \dfrac{\sqrt{2}-1}{2\sqrt{2}} \right)}^{\dfrac{1}{2}}}$$ Rationalizing the denominator by multiplying and dividing with $$\sqrt{2}$$, we get, $$\begin{aligned} & \Rightarrow \sin \theta ={{\left( \dfrac{\sqrt{2}-1}{2\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}} \right)}^{\dfrac{1}{2}}} \\\ & \Rightarrow \sin \theta ={{\left( \dfrac{2-\sqrt{2}}{4} \right)}^{\dfrac{1}{2}}} \\\ & \Rightarrow \sin \theta =\dfrac{{{\left( 2-\sqrt{2} \right)}^{\dfrac{1}{2}}}}{2} \\\ & \Rightarrow \sin \left( {{22}^{\circ }}30' \right)=\dfrac{{{\left( 2-\sqrt{2} \right)}^{\dfrac{1}{2}}}}{2} \\\ \end{aligned}$$ Hence, the value of given trigonometric function is $$\dfrac{{{\left( 2-\sqrt{2} \right)}^{\dfrac{1}{2}}}}{2}$$. **Note:** One may note that in trigonometry the term ‘minute’ denotes the angle and not time. Always remember that: - $$60'={{1}^{\circ }}$$ and $$60''={{1}^{\circ }}$$ where 60’’ represents 60 seconds. It will be very difficult for us to determine the value of $$\sin \left( {{22.5}^{\circ }} \right)$$ without using the half angle formula, so you must remember these basic formulas like: - $${{\cos }^{2}}\theta =\dfrac{1+\cos 2\theta }{2},{{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}$$ and $$\sin 2\theta =2\sin \theta \cos \theta $$. Note that here we have used the second formula. You can also use the first one to get the answer.