# trigonometry

if x be real,prove that
x^2 -2xcosA +1
x^2 -2xcosB +1
lies between
sin^2(A/2)/sin^2(B/2) AND cos^2(A/2)/cos^2(B/2).

\hspace{-16}$Given$\bf{y=\frac{x^2-2x\cos A+1}{x^2-2x\cos B+1} = \frac{\left(x^2-2x\cos B+1\right)+\left(2x\cos B-2x\cos A\right)}{x^2-2x\cos B+1},}$\\\\\\$\bf{y=1+\frac{2x\cdot \left(\cos B-\cos A\right)}{x^2-2x\cos B+1}=1+\frac{2\left(\cos B-\cos A\right)}{\left(x^2+\frac{1}{x^2}\right)-2\cos B}}\;\; $, where$\bf{x\in \mathbb{R}-\{0\}}$\\\\\\ Now we will find Range of$\bf{\left(x^2+\frac{1}{x^2}\right)}$for$\bf{x\in \mathbb{R}-\{0\}}$\\\\\\$\bullet $If$\bf{x>0}$, Then$\bf{\left(x^2+\frac{1}{x^2}\right)\geq 2}$using$\bf{A.M\geq G.M}$\\\\\\ So$\bf{y_{Max.} = 1+\frac{2\left(\cos B-\cos A\right)}{2-2\cos B} = \frac{1-\cos A}{1-\cos B}=\frac{\sin ^2 \frac{A}{2}}{\sin ^2 \frac{B}{2}}}$\\\\\\$\bullet $If$\bf{x<0}$, Then$\bf{\left(x^2+\frac{1}{x^2}\right)\leq -2}$using$\bf{A.M\geq G.M}$\\\\\\ So$\bf{y_{Min.} = 1+\frac{2\left(\cos B-\cos A\right)}{-2-2\cos B} = \frac{1+\cos A}{1+\cos B}=\frac{\cos ^2 \frac{A}{2}}{\cos ^2 \frac{B}{2}}}$\\\\\\ \hspace{-16}$So $\bf{\frac{\cos ^2 \frac{A}{2}}{\cos ^2 \frac{B}{2}} \leq y \leq \frac{\sin ^2 \frac{A}{2}}{\sin ^2 \frac{B}{2}}}$\\\\\\ and at $\bf{x=0}$, we get $\bf{y=1}$,\\\\\\ which is possible when $\bf{A = B}$