Question

If sin∅=3/5 and ∅ is acute find sin½∅

Answer 1

Answer to Question #88535 – Math – Trigonometry

Question

If $\sin \varnothing = 3/5$ and $\varnothing$ is acute find $\sin \frac{1}{2}\varnothing$

Solution

$\sin \varnothing = \frac{3}{5}$

We know that $\sin^2 \varnothing + \cos^2 \varnothing = 1 \Rightarrow \cos^2 \varnothing = 1 - \sin^2 \varnothing$

$\cos \varnothing = \sqrt{1 - \sin^2 \varnothing}$ (By taking square root on both sides)

$$\begin{array}{l} = \sqrt{1 - \left(\frac{3}{5}\right)^2} \text{ (by substituting } \sin \varnothing = \frac{3}{5}) \\ = \sqrt{1 - \frac{9}{25}} \\ = \sqrt{\frac{25 - 9}{25}} \\ = \sqrt{\frac{16}{25}} \\ = \sqrt{\left(\frac{4}{5}\right)^2} \\ = \frac{4}{5} \text{ (Since } \sqrt{x^2} = x) \end{array}$$

Therefore, $\cos \varnothing = \frac{4}{5}$.

We know that $\varnothing$ is an acute angle, then $\varnothing/2$ also will be an acute angle. Besides,

$$\begin{array}{l} \cos \varnothing = 1 - 2 \sin^2 \left(\frac{\varnothing}{2}\right) \\ \Rightarrow 2 \sin^2 \left(\frac{\varnothing}{2}\right) = 1 - \cos \varnothing \\ \Rightarrow \sin^2 \left(\frac{\varnothing}{2}\right) = \frac{1 - \cos \varnothing}{2} \text{ (by dividing both sides by 2)} \\ \Rightarrow \sin \left(\frac{\varnothing}{2}\right) = \sqrt{\frac{1 - \cos \varnothing}{2}} \text{ (by taking square root on both sides)} \\ = \sqrt{\frac{1 - \frac{4}{5}}{2}} \text{ (by substituting } \cos \varnothing = \frac{4}{5}) \\ = \sqrt{\frac{\frac{5 - 4}{5}}{2}} \\ = \sqrt{\frac{\left(\frac{1}{5}\right)}{2}} = \sqrt{\frac{1}{10}} \end{array}$$

Therefore, $\sin \left(\frac{\varnothing}{2}\right) = \frac{1}{\sqrt{10}} = 0.3162$.

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