Question
$4\Big(\text{bc}\cos^2\frac{\text{A}}{2}+\text{ca}\cos^2\frac{\text{B}}{2}+\text{ab}\cos^2\frac{\text{C}}{2}\Big)=(\text{a + b + c})^2$
$\text{LHS}=4\Big(\text{bc}\cos^2\frac{\text{A}}{2}+\text{ca}\cos^2\frac{\text{B}}{2}+\text{ab}\cos^2\frac{\text{C}}{2}\Big)$
$=2\Big(\text{bc.}2\cos^2\frac{\text{A}}{2}+\text{ca.}2\cos^2\frac{\text{B}}{2}+\text{ab}.2\cos^2\frac{\text{C}}{2}\Big)$
$=2(\text{bc.}(1-\cos\text{A})+\text{ca.}(1-\cos\text{B})+\text{ab.}(1-\cos\text{C}))$
$=2\text{bc}-2\text{bc}\cos\text{A}+2\text{ca}-2\text{ca}\cos\text{B}+2\text{ab}-2\text{ab}\cos\text{C}$
$=2\text{bc}-2\text{bc}\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}+2\text{ca}-2\text{ca}\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ca}}+2\text{ab}$
$-2\text{ab}\frac{\text{b}^2+\text{a}^2-\text{c}^2}{2\text{ab}}\text{[cos rule]}$
$=2\text{bc}-\text{b}^2-\text{c}^2+\text{a}^2+2\text{ca}-\text{a}^2-\text{c}^2+\text{b}^2+2\text{ab}-\text{b}^2-\text{a}^2+\text{c}^2$
$=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{ab}+2\text{ca}$
$=(\text{a + b + c})^2=\text{RHS}$
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| (i) | $((\text{A}'\cup\text{B}')-\text{A})'$ | (a) | $\text{A} - \text{B}$ |
| (ii) | $[\text{B}'\cup(\text{B}'-\text{A})]'$ | (b) | $\text{A}$ |
| (iii) | $(\text{A} - \text{B}) - (\text{B} - \text{C})$ | (c) | $\text{B}$ |
| (iv) | $(\text{A}-\text{B})\cap(\text{C}-\text{B})$ | (d) | $(\text{A}\times\text{B})\cap(\text{A}\times\text{C})$ |
| (v) | $\text{A}\times(\text{B}\cap\text{C})$ | (e) | $(\text{A}\times\text{B})\cup(\text{A}\times\text{C})$ |
| (vi) | $\text{A}\times(\text{B}\cup\text{C})$ | (f) | $(\text{A}\cap\text{C})-\text{B}$ |