Proof for an inequality(Trigonometric functions) I would appreciate if somebody could help me with the following problem
Q: How to proof 
$$\sqrt{1+a^2-2a\cos t}\leq (1-a)+2a\sin\frac{t}{2}(\text{where}~~ 0\leq a\leq 1, 0\leq t\leq 2\pi)$$

Putting $t=2u,$
$$1+a^2-2a\cos2u=1+a^2-2a(1-2\sin^2u)=(1-a)^2+a(2\sin u)^2$$ which will be $$\le (1-a)^2+a^2(2\sin u)^2+2(1-a)a(2\sin u)$$
$$\text{iff  }a(2\sin u)^2\le a^2(2\sin u)^2+2(1-a)a(2\sin u) $$
$$\text{iff  }a\sin u\le a^2\sin u+(1-a)a \text{ as }\sin u \ge0\text{ as }0\le u\le \pi$$
$$\text{iff  }a(1-a)\sin u\le (1-a)a $$
$$\text{iff  }\sin u\le 1 \text{ as } 0\le a(1-a)\le 1 \text{ as } 0\le a\le1$$
which is true
$$\implies 1+a^2-2a\cos2u\le (1-a+2a\sin u)^2$$  
Now, follow the advice by GitGud
