If $ \sin (A-B)=\sin A \cos B-\cos A \sin B $ and $ \cos (A-B)=\cos A \cos B+\sin A \sin B $, find the values of $ \sin 15^{\circ} $ and $ \cos 15^{\circ} $.

Given:

\( \sin (A-B)=\sin A \cos B-\cos A \sin B \) and \( \cos (A-B)=\cos A \cos B+\sin A \sin B \).

To do:

We have to find the values of \( \sin 15^{\circ} \) and \( \cos 15^{\circ} \).

Solution:  

Let $A=45^{\circ}, B=30^{\circ}$

We know that,

$\sin 45^{\circ}=\frac{1}{\sqrt2}$

$\sin 30^{\circ}=\frac{1}{2}$

$\cos 30^{\circ}=\frac{\sqrt3}{2}$

$\cos 45^{\circ}=\frac{1}{\sqrt2}$

Therefore,

$\sin (\mathrm{A}-\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}-\cos \mathrm{A} \sin \mathrm{B}$

$\Rightarrow  \sin \left(45^{\circ}-30^{\circ}\right)=\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ}$

$\Rightarrow \sin 15^{\circ}=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}$

$=\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}$

$=\frac{\sqrt{3}-1}{2 \sqrt{2}}$

$\cos (\mathrm{A}-\mathrm{B})=\cos \mathrm{A} \cos \mathrm{B}+\sin \mathrm{A} \sin \mathrm{B}$

$\Rightarrow \cos \left(45^{\circ}-30^{\circ}\right)=\cos 45^{\circ} \cos 30^{\circ}+\sin .45^{\circ} \sin 30^{\circ}$

$\Rightarrow \cos 15^{\circ}=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}$

$=\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}$

$=\frac{\sqrt{3}+1}{2 \sqrt{2}}$

The values of \( \sin 15^{\circ} \) and \( \cos 15^{\circ} \) are $\frac{\sqrt{3}-1}{2 \sqrt{2}}$ and $\frac{\sqrt{3}+1}{2 \sqrt{2}}$ respectively.

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Updated on: 10-Oct-2022

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