Question

How do you find the values of How do you find the values of $\sin {{22.5}^{\circ }}$using half angle formula?

Answer 1

In this question we will take $\theta ={{22.5}^{\circ }}$. From that we will find out $2\theta$ so we get $2\theta ={{45}^{\circ }}$.
$cos2\theta =1-2{{\sin }^{2}}\theta$is the basic trigonometric formula used to solve this question. We have to know the value $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$.

Complete step by step answer:
From the question we were asked to find the values of $\sin {{22.5}^{\circ }}$using the half angle formula.
Let us solve this question by taking $\theta ={{22.5}^{\circ }}$
$\Rightarrow \theta ={{22.5}^{\circ }}$
Now multiply with $2$on both sides, so we get
$\Rightarrow 2\times \theta =2\times {{22.5}^{\circ }}$
$\Rightarrow 2\theta ={{45}^{\circ }}$
Now consider the basic trigonometric formula$cos2\theta =1-2{{\sin }^{2}}\theta$
$\Rightarrow cos2\theta =1-2{{\sin }^{2}}\theta$…………………(1)
Using transformations, this formula can be modified into
$\Rightarrow {{\sin }^{2}}\theta =\dfrac{1}{2}\left( 1-\cos 2\theta \right)$………………..(2)
Now put $\theta ={{22.5}^{\circ }}$ in equation (2), we get
$\Rightarrow {{\sin }^{2}}\theta =\dfrac{1}{2}\left( 1-\cos 2\theta \right)$
$\Rightarrow {{\sin }^{2}}{{22.5}^{\circ }}=\dfrac{1}{2}\left( 1-\cos (2\times {{22.5}^{\circ }}) \right)$
We know that $2\times {{22.5}^{\circ }}={{45}^{\circ }}$
$\Rightarrow {{\sin }^{2}}{{22.5}^{\circ }}=\dfrac{1}{2}\left( 1-\cos ({{45}^{\circ }}) \right)$…………..(3)
Since we know $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$, we can put $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$in equation(3)
$\Rightarrow {{\sin }^{2}}{{22.5}^{\circ }}=\dfrac{1}{2}\left( 1-\dfrac{1}{\sqrt{2}} \right)$
After simplification we get,
$\Rightarrow {{\sin }^{2}}{{22.5}^{\circ }}=\dfrac{1}{2}\left( \dfrac{\sqrt{2}-1}{\sqrt{2}} \right)$
$\Rightarrow {{\sin }^{2}}{{22.5}^{\circ }}=\dfrac{\sqrt{2}-1}{2\sqrt{2}}$
$\Rightarrow {{\sin }^{2}}{{22.5}^{\circ }}=0.14639$
Apply square root on both sides, we get
$\Rightarrow \sqrt{{{\sin }^{2}}{{22.5}^{\circ }}}=\sqrt{0.14639}$
$\Rightarrow \sqrt{{{\sin }^{2}}{{22.5}^{\circ }}}=\pm \sqrt{0.14639}$
$\Rightarrow \sin {{22.5}^{\circ }}=\pm 0.3826$
Since ${{22.5}^{\circ }}$ is in the first quadrant, we will take $\sin {{22.5}^{\circ }}$ as positive.
So, we take only $+0.3826$
$\Rightarrow \sin {{22.5}^{\circ }}=+0.3826$
So finally, we can conclude that the value of $\sin {{22.5}^{\circ }}=+0.3826$.

Note: students should be careful while doing calculations. Because small calculation errors can make the final answer wrong. Students should apply the correct formulas to get answers very quickly and easily. Many students may have misconception that $+0.3826$and $-0.3826$ are both values are right but actually $+0.3826$ is correct answer because ${{22.5}^{\circ }}$is in first quadrant, in first quadrant $\sin {{22.5}^{\circ }}$ as positive value.