Sure, let's prove that [tex]\(\cos 15^\circ - \sin 15^\circ = \frac{1}{\sqrt{2}}\)[/tex]. We will proceed step-by-step:
1. Evaluate [tex]\(\cos 15^\circ\)[/tex]:
[tex]\[
\cos 15^\circ = \cos (45^\circ - 30^\circ)
\][/tex]
Using the cosine difference identity [tex]\(\cos (a - b) = \cos a \cos b + \sin a \sin b\)[/tex], we have:
[tex]\[
\cos 15^\circ = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ
\][/tex]
Substituting the known values [tex]\(\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}}\)[/tex], [tex]\(\cos 30^\circ = \frac{\sqrt{3}}{2}\)[/tex], and [tex]\(\sin 30^\circ = \frac{1}{2}\)[/tex], we get:
[tex]\[
\cos 15^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right)
\][/tex]
[tex]\[
\cos 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}
\][/tex]
2. Evaluate [tex]\(\sin 15^\circ\)[/tex]:
[tex]\[
\sin 15^\circ = \sin (45^\circ - 30^\circ)
\][/tex]
Using the sine difference identity [tex]\(\sin (a - b) = \sin a \cos b - \cos a \sin b\)[/tex], we have:
[tex]\[
\sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ
\][/tex]
Again, substituting the known values:
[tex]\[
\sin 15^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right)
\][/tex]
[tex]\[
\sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}}
\][/tex]
3. Calculate [tex]\(\cos 15^\circ - \sin 15^\circ\)[/tex]:
[tex]\[
\cos 15^\circ - \sin 15^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}} - \frac{\sqrt{3} - 1}{2\sqrt{2}}
\][/tex]
[tex]\[
\cos 15^\circ - \sin 15^\circ = \frac{\sqrt{3} + 1 - (\sqrt{3} - 1)}{2\sqrt{2}}
\][/tex]
[tex]\[
\cos 15^\circ - \sin 15^\circ = \frac{\sqrt{3} + 1 - \sqrt{3} + 1}{2\sqrt{2}}
\][/tex]
[tex]\[
\cos 15^\circ - \sin 15^\circ = \frac{2}{2\sqrt{2}}
\][/tex]
[tex]\[
\cos 15^\circ - \sin 15^\circ = \frac{1}{\sqrt{2}}
\][/tex]
Therefore, we have proven that:
[tex]\[
\cos 15^\circ - \sin 15^\circ = \frac{1}{\sqrt{2}}
\][/tex]
The numerical values confirm this as [tex]\(\cos 15^\circ \approx 0.9659\)[/tex], [tex]\(\sin 15^\circ \approx 0.2588\)[/tex], and their difference [tex]\(\approx 0.7071\)[/tex], which is indeed [tex]\(\frac{1}{\sqrt{2}}\)[/tex]. Finally, this matches the theoretical verification.
