Prove that: \[ \sin 105^\circ + \cos 105^\circ = \cos 45^\circ \]
Solution
Using the identity:
\[
\sin \theta + \cos \theta
=
\sqrt{2}\sin(\theta+45^\circ)
\]
Taking
\[
\theta = 105^\circ
\]
Then,
\[
\sin 105^\circ + \cos 105^\circ
=
\sqrt{2}\sin(105^\circ+45^\circ)
\]
\[
=
\sqrt{2}\sin 150^\circ
\]
\[
=
\sqrt{2}\times\frac{1}{2}
\]
\[
=
\frac{1}{\sqrt{2}}
\]
Now,
\[
\cos 45^\circ
=
\frac{1}{\sqrt{2}}
\]
Therefore,
\[
\sin 105^\circ + \cos 105^\circ
=
\cos 45^\circ
\]
Hence,
\[
\boxed{
\sin 105^\circ + \cos 105^\circ
=
\cos 45^\circ
}
\]