Question

Find the value of $\sin {15}^{\circ }$using $\sin {30}^{\circ }$.

Answer 1

Step 1: Write the first trigonometric function in the form of the second trigonometric function :

Given trigonometric functions: $\sin {15}^{\circ }$ and $\sin {30}^{\circ }$6282a90fcb5fe67a7b1b9ada 6282a90fcb5fe67a7b1b9ada

We know that for all values of the angle $A$,

$$\begin{gathered} {\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right)}^2={\sin }^2\frac{A}{2}+\cos {}^2\frac{A}{2}+2\sin \frac{A}{2}\cos \frac{A}{2} \\ =1+\sin (2\times \frac{A}{2})[\because {\sin }^2\theta +{\cos }^2\theta =1and\sin 2\theta =2\sin \theta cos\theta ] \\ =1+\sin A \end{gathered}$$

Therefore, $\sin \frac{A}{2}+\cos \frac{A}{2}=\pm \sqrt{1+\sin \frac{A}{2}}....\left(i\right)$

Now, let $A=30°$ then, $\frac{A}{2}=\frac{30°}{2}=15°$ and from the above equation we get,
$\sin {15}^{\circ }+\cos {15}^{\circ }=\pm \sqrt{1+\sin {30}^{\circ }}$

Similarly, for all values of the angle $A$,

$$\begin{gathered} {\left(\sin \frac{A}{2}-\cos \frac{A}{2}\right)}^2={\sin }^2\frac{A}{2}+\cos {}^2\frac{A}{2}-2\sin \frac{A}{2}\cos \frac{A}{2} \\ =1-\sin (2\times \frac{A}{2})[\because {\sin }^2\theta +{\cos }^2\theta =1and\sin 2\theta =2\sin \theta cos\theta ] \\ =1-\sin A \end{gathered}$$

Therefore, $\sin \frac{A}{2}-\cos \frac{A}{2}=\pm \sqrt{1+\sin \frac{A}{2}}...\left(ii\right)$

Now, let $A=30°$ then, $\frac{A}{2}=\frac{30°}{2}=15°$ and from the above equation, we get
$\sin {15}^{\circ }-\cos {15}^{\circ }=\pm \sqrt{1+\sin {30}^{\circ }}$

Step 2: Find the exact values.

Clearly, $\sin 15°>0$and $\cos 15˚>0$

Therefore, $\sin 15°+\cos 15°>0$

Therefore, from equation $\left(i\right)$ we get,
$$\begin{gathered} \sin {15}^{\circ }+\cos {15}^{\circ }=+\sqrt{1+\sin {30}^{\circ }} \\ \Rightarrow \sin {15}^{\circ }+\cos {15}^{\circ }=\sqrt{1+\frac{1}{2}}...\left(iii\right) \end{gathered}$$

Again,

$$\begin{gathered} \sin {15}^{\circ }-\cos {15}^{\circ }=\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin {15}^{\circ }-\frac{1}{\sqrt{2}}\cos {15}^{\circ }\right) \\ =\sqrt{2}\left(\cos 45°\sin {15}^{\circ }-\sin 45°\cos {15}^{\circ }\right) \\ =\sqrt{2}\sin ({15}^{\circ }–{45}^{\circ }) \\ =\sqrt{2}\sin \left(-{30}^{\circ }\right) \\ =-\sqrt{2}\sin 30° \\ =-\sqrt{2}\times \frac{1}{2} \\ =-\frac{1}{\sqrt{2}} \end{gathered}$$

Thus, $\sin 15°-\cos 15°<0$

Therefore, from equation $\left(ii\right)$ we get,

$$\begin{gathered} \sin {15}^{\circ }-\cos {15}^{\circ }=-\sqrt{1+\sin {30}^{\circ }} \\ \Rightarrow \sin {15}^{\circ }-\cos {15}^{\circ }=-\sqrt{1+\frac{1}{2}}...\left(iv\right) \end{gathered}$$

Step 3: Find the value of the required trigonometric function.

Now, adding equation $\left(iii\right)$ and equation $\left(iv\right)$ we get,

$$\begin{gathered} 2\sin {15}^{\circ }=\sqrt{1+\frac{1}{2}}-\sqrt{1-\frac{1}{2}} \\ \Rightarrow 2\sin {15}^{\circ }=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}} \\ \Rightarrow 2\sin {15}^{\circ }=\frac{\sqrt{3}-1}{\sqrt{2}} \\ \Rightarrow \sin {15}^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}} \end{gathered}$$

Hence, the value of $\sin {15}^{\circ }$is$\frac{\sqrt{3}-1}{2\sqrt{2}}$.