X=cos\alp,\alp∈[0:;π] \displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\L eftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+
1−x
2


∣=
2

(2x
2
−1)\Leftright∣cos\alp+sin\alp∣=
2

(2cos
2
\alp−1)
\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|= \N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N
2

cos(\alp−
4
π

)∣=N
2

cos(2\alp)\Right\alp∈[0;
4
π

]∪[
4
3π

;π]
1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;

PLEASE SIMPLIFY