If sin (A − B) = sin A cos B − cos A sin B and cos (A − B) = cos A cos B + sin A sin B,

Question:

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 $\cos 15^{\circ}$.

Solution:

Given:

$\sin (A-B)=\sin A \cos B-\cos A \sin B \ldots \ldots(1)$

$\cos (A-B)=\cos A \cos B+\sin A \sin B$....(2)

To find:

The values of $\sin 15^{\circ}$ and $\cos 15^{\circ}$

In this problem we need to find $\sin 15^{\circ}$ and $\cos 15^{\circ}$

Hence to get $15^{\circ}$ angle we need to choose the value of $A$ and $B$ such that $(A-B)=15^{\circ}$

So If we choose $A=45^{\circ}$ and $B=30^{\circ}$

Then we get, $(A-B)=15^{\circ}$

Therefore by substituting $A=45^{\circ}$ and $B=30^{\circ}$ in equation (1)

We get,

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

Therefore,

$\sin \left(15^{\circ}\right)=\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ}$....(3)

Now we know that,

$\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \sin 30^{\circ}=\frac{1}{2}, \cos 30^{\circ}=\frac{\sqrt{3}}{2}$

Now by substituting above values in equation (3)

We get,

$\sin \left(15^{\circ}\right)=\left(\frac{1}{\sqrt{2}}\right) \times\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{\sqrt{2}}\right) \times\left(\frac{1}{2}\right)$

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

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

Therefore,

$\sin \left(15^{\circ}\right)=\frac{\sqrt{3}-1}{2 \sqrt{2}} \ldots \ldots$(4)

Now by substituting $A=45^{\circ}$ and $B=30^{\circ}$ in equation (2)

We get,

$\cos \left(45^{\circ}-30^{\circ}\right)=\cos 45^{\circ} \cos 30^{\circ}+\sin 45^{\circ} \sin 30^{\circ}$....(2)

Now we know that,

$\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \sin 30^{\circ}=\frac{1}{2}, \cos 30^{\circ}=\frac{\sqrt{3}}{2}$

Now by substituting above values in equation (5)

We get,

$\cos \left(15^{\circ}\right)=\left(\frac{1}{\sqrt{2}}\right) \times\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\right) \times\left(\frac{1}{2}\right)$

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

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

Therefore,

$\cos \left(15^{\circ}\right)=\frac{\sqrt{3}+1}{2 \sqrt{2}} \ldots \ldots$(6)

Therefore from equation (4) and (6)

$\sin \left(15^{\circ}\right)=\frac{\sqrt{3}-1}{2 \sqrt{2}}$

$\cos \left(15^{\circ}\right)=\frac{\sqrt{3}+1}{2 \sqrt{2}}$