## Solution to another S L Loney problem for IITJEE Advanced Mathematics

Exercise XXVII. Problem #32. The sides of a triangle are in AP and the greatest and least angles are $\theta$ and $\phi$. Prove that

$4(1-\cos{\theta})(1-\cos{\phi})=\cos{\theta}+\cos{\phi}$

Proof:

Let a, b, c be in AP. Hence, $a, which in turn implies c is greatest and a is the least.

Hence, $\theta=\angle{C}$ and $\angle{A}$.

Want:

$4(1-\cos{A})(1-\cos{C})=\cos{A}+\cos{C}$

Given:

$b-a=c-b$

$(b-a)^{2}=(b-c)^{2}$. Hence,

$b^{2}-2ab+a^{2}=b^{2}+c^{2}-2bc$

$b^{2}+a^{2}-c^{2}=b^{2}-2bc+2ab$

$2b=a+c$

$4b^{2}=a^{2}+c^{2}+2ac$

$a^{2}+c^{2}-b^{2}=3b^{2}-2ac$

$a^{2}+b^{2}-c^{2}=b^{2}-2bc+2ab$

$b^{2}+c^{2}-a^{2}=b^{2}-2ab+2bc$

Now, we have $1-\cos{A}=1-\frac{b^{2}+c^{2}-a^{2}}{2bc}$ and this is equal to the following:

$\frac{2bc-(b^{2}+c^{2}-a^{2})}{2bc}=\frac{2bc-(b^{2}-2ab+2bc)}{2bc}$

$= \frac{2bc-b^{2}+2ab-2bc}{2bc}=\frac{2ab-b^{2}}{2bc}$.

Also, similarly, we have the following:

$1-\cos{C}=1-\frac{a^{2}+b^{2}-c^{2}}{2ab}=1-\frac{b^{2}-2bc+2ab}{2ab}=\frac{2bc-b^{2}}{2ab}$

$\cos{A}+\cos{C}=\frac{b^{2}+c^{2}-a^{2}}{}+\frac{a^{2}+b^{2}-c^{2}}{2ab}$

$= \frac{b^{2}-2ab+2bc}{2bc}+\frac{b^{2}-2bc+2ab}{2ab}$

$=\frac{a(b^{2}-2ab+2bc)+c(b^{2}-2bc+2ab)}{2abc}$

$=\frac{ab^{2}-2a^{2}b+2abc+b^{2}c-2bc^{2}+2abc}{2abc}$

$=\frac{ab^{2}+b^{2}c-2a^{2}b-2c^{2}b+4abc}{2abc}$

$= \frac{ab+bc-2a^{2}-2c^{2}+4ac}{2ac}$

$=\frac{b(a+c)-2(a-c)^{2}}{2ac}$

$=\frac{2b^{2}-2(a-c)^{2}}{2ac}$

$=\frac{b^{2}-(a-c)^{2}}{2ac}$

$\frac{(b+a-c)}{(b-a+c)}{ac}$

but it is given that $b-a=c-b$, hence, $a+c=2b$, $a+c-b=b$. So the above expression changes to

$= \frac{(b+a-c)(c-b+c)}{ac}$

$= \frac{(2c-b)(a+b-c)}{ac}$

$= \frac{(2c-b)(a+b-\overline{2b-a})}{ac}$

$= \frac{(2c-b)(a-b+c)}{ac}$

$= \frac{(2c-b)(2a-b)}{ac}$

$= \frac{4ac-2bc-2ab+b^{2}}{ac}$

$= 4(1-\cos{A})(1-\cos{C})$.

QED.

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