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If $\cos^{-1}x - \cos^{-1}\!\dfrac{y}{2}=\alpha$ where $-1\leq x\leq1,\ -2\leq y\leq2,\ x\leq\dfrac{y}{2}$, then $4x^2-4xy\cos\alpha+y^2$ equals:
Let $\cos^{-1}x=A$ and $\cos^{-1}(y/2)=B$, so $A-B=\alpha$.
Then $\cos A=x,\ \cos B=y/2,\ \sin A=\sqrt{1-x^2},\ \sin B=\sqrt{1-y^2/4}$.
Using $\cos\alpha=\cos(A-B)=\cos A\cos B+\sin A\sin B=\dfrac{xy}{2}+\sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}}$
$4x^2-4xy\cos\alpha+y^2 = (2x-y\cos\alpha)^2+y^2(1-\cos^2\alpha)$... expanding directly: $= 4x^2+y^2-4xy\cos\alpha$.
After substitution and simplification using the expression for $\cos\alpha$: $= 4(1-\cos^2\alpha) = \mathbf{4\sin^2\alpha}$.
Bulk Modulus $K = \frac{\Delta P}{-\Delta V/V}$.
- The change in pressure is $\Delta P = \frac{mg}{a}$.
- The fractional change in volume for a sphere is $\frac{\Delta V}{V} = 3\frac{dr}{r}$.
- $K = \frac{mg/a}{3dr/r} \implies \frac{dr}{r} = \frac{mg}{3Ka}$.
Since quantities added or subtracted must have the same dimensions, $\dfrac{a}{V^2}$ must have the same dimensions as pressure $P$.
$[P] = [ML^{-1}T^{-2}]$ and $[V^2] = [L^6]$
Therefore: $[a] = [P][V^2] = [ML^{-1}T^{-2}][L^6] = [ML^5T^{-2}]$
Acetic acid can be prepared by hydrolysis of:
Hydrolysis of methyl cyanide (acetonitrile, CH₃CN) yields acetic acid: \(\mathrm{CH_3CN + 2H_2O \rightarrow CH_3COOH + NH_3}\).
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