Use the identity \(\cos 2\theta  = 2{\cos ^2}\theta  - 1\) to prove that \(\cos \frac{1}{2}x = \sqrt {\frac{{1 + \cos x}}{2}} ,{\text{ }}0 \leqslant x \leqslant \pi \).

Find a similar expression for \(\sin \frac{1}{2}x,{\text{ }}0 \leqslant x \leqslant \pi \).

Hence find the value of \(\int_0^{\frac{\pi }{2}} {\left( {\sqrt {1 + \cos x}  + \sqrt {1 - \cos x} } \right){\text{d}}x} \).

\(\cos x = 2{\cos ^2}\frac{1}{2}x - 1\)

\(\cos \frac{1}{2}x =  \pm \sqrt {\frac{{1 + \cos x}}{2}} \)     M1

positive as \(0 \leqslant x \leqslant \pi \)     R1

\(\cos \frac{1}{2}x = \sqrt {\frac{{1 + \cos x}}{2}} \)     AG

[2 marks]

\(\cos 2\theta  = 1 - 2{\sin ^2}\theta \)     (M1)

\(\sin \frac{1}{2}x = \sqrt {\frac{{1 - \cos x}}{2}} \)     A1

[2 marks]

\(\sqrt 2 \int_0^{\frac{\pi }{2}} {\cos \frac{1}{2}x + \sin \frac{1}{2}x{\text{d}}x} \)     A1

\( = \sqrt 2 \left[ {2\sin \frac{1}{2}x - 2\cos \frac{1}{2}x} \right]_0^{\frac{\pi }{2}}\)     A1

\( = \sqrt 2 (0) - \sqrt 2 (0 - 2)\)     A1

\( = 2\sqrt 2 \)     A1

[4 marks]